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17–1. z

Determine the moment of inertia Iy for the slender rod. The rod’s density r and cross-sectional area A are constant.

Express the result in terms of the rod’s total mass m.

y

l

SOLUTION

A x

Iy =

LM

x 2 dm

l

= =

L0

x 2 (r A dx)

1 r A l3 3

m = rAl Thus, Iy =

1 m l2 3

Ans.

Ans: Iy = 791

1 m l2 3

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17–2. z

The solid cylinder has an outer radius R, height h, and is made from a material having a density that varies from its center as r = k + ar2, where k and a are constants.

Determine the mass of the cylinder and its moment of inertia about the z axis.

R

h

SOLUTION Consider a shell element of radius r and mass dm = r dV = r(2p r dr)h R

m =

(k + ar2)(2p r dr)h

L0

m = 2p h(

aR4 kR2 + ) 2 4

m = p h R2(k +

aR2 ) 2

Ans.

dI = r2 dm = r2(r)(2p r dr)h R

Iz =

L0

r2(k + ar2)(2p r dr) h R

Iz = 2ph

L0

Iz = 2ph[ Iz =

(k r3 + a r5) dr

k R4 aR6 + ] 4 6

2 aR2 p h R4 [k + ] 2 3

Ans.

Ans:

aR2 b 2 p h R4 2 aR2 Iz = ck + d 2 3

m = p h R2 ak +

792

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17–3. y

Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.

R x

SOLUTION 2p

Iz =

L0

r A(R du)R2 = 2p r A R3 2p

m =

L0

r A R du = 2p r A R

Thus, I z = m R2

Ans.

Ans: Iz = mR2 793

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*17–4. y

The paraboloid is formed by revolving the shaded area around the x axis. Determine the radius of gyration kx. The density of the material is r = 5 Mg>m3.

y2

50x 100 mm x

SOLUTION 200 mm

dm = r p y2 dx = r p (50x) dx Ix =

1 2 1 y dm = 2 L0 L2

= r pa

200

50 x {p r (50x)} dx

502 1 3 200 bc x d 2 3 0

= rp a

502 b(200)3 6 200

m =

L

dm =

L0

p r (50x) dx

200 1 = r p (50)c x2 d 2 0

= rp a

kx =

50 b (200)2 2

Ix 50 (200) = 57.7 mm = Am A3

Ans.

Ans: kx = 57.7 mm 794

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17–5. y

Determine the radius of gyration kx of the body. The specific weight of the material is g = 380 lb>ft3.

y3 = x 2 in. x

8 in.

SOLUTION dm = r dV = rp y2 dx d Ix =

1 1 (dm) y2 = pry4 dx 2 2 8

Ix =

1

L0 2

prx4/3 dx = 86.17r

8

m = kx =

L0

prx2/3 dx = 60.32r

86.17r Ix = = 1.20 in.

Am A 60.32r

Ans.

Ans: kx = 1.20 in.

795

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17–6. y

The sphere is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the sphere. The material has a constant density r.

x2 + y2 = r2

x

SOLUTION dIx =

y2 dm 2

dm = r dV = r(py2 dx) = r p(r2 - x2) dx dIx =

1 r p(r2 - x2)2 dx 2 r

Ix = =

1 2 2 2 r p(r - x ) dx 2 L- r 8 pr r5 15 r

m = =

L- r

2

2

Ix =

2 m r2 5

r p(r - x ) dx

4 r p r3 3

Thus, Ans.

Ans: Ix = 796

2 mr2 5

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17–7. y

The frustum is formed by rotating the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the frustum. The frustum has a constant density r.

y

–ba x

b 2b

b

x z

SOLUTION

a

2b2 b2 x + b2 B dx dm = r dV = rpy2 dx = rp A 2 x2 + a a dIx =

1 1 dmy2 = rpy4 dx 2 2

dIx =

1 4 b4 6 b4 4b4 b4 rp A 4 x4 + 3 x3 + 2 x2 + x + b4 B dx a 2 a a a a

Ix =

L

dIx = =

1 4b4 6 b4 4 b4 b4 x + b4 B dx rp A 4 x4 + 3 x3 + 2 x2 + a 2 L0 a a a 31 rpab4 10 a

m = Ix =

Lm

dm = rp

L0

A 2 x2 + b2 a

2b2 7 x + b2 B dx = rpab2 a 3

93 2 mb 70

Ans.

Ans: Ix = 797

93 2 mb 70

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*17–8. y

The hemisphere is formed by rotating the shaded area around the y axis. Determine the moment of inertia Iy and express the result in terms of the total mass m of the hemisphere. The material has a constant density r.

x2

y2

r2

x

SOLUTION r

m =

LV

r dV = r

= rp cr 2 y -

Iy = =

L0

r

p x2 dy = rp

L0

(r2 - y2)dy

1 3 r 2 y d = rp r3 3 3 0

r r r rp 1 (dm) x2 = px4 dy = (r2 - y2)2 dy 2 L0 2 L0 Lm 2

rp 4 y5 r 4rp 5 2 c r y - r2 y3 + d = r 2 3 5 0 15

Thus, Iy =

2 m r2 5

Ans.

Ans: Iy = 798

2 mr 2 5

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17–9.

Determine the moment of inertia of the homogeneous triangular prism with respect to the y axis. Express the result in terms of the mass m of the prism. Hint: For integration, use thin plate elements parallel to the x–y plane and having a thickness dz.

z

–h (x – a) z = –– a

h

SOLUTION dV = bx dz = b(a)(1 -

z ) dz h

x

b

a

y

x dIy = dIy + (dm)[( )2 + z2] 2 =

x2 1 dm(x2) + dm( ) + dmz2 12 4

= dm(

x2 + z2) 3

= [b(a)(1 -

a2 z z )dz](r)[ (1 - )2 + z2] h 3 h

k

Iy = abr

a3 h - z 3 z ( ) + z2(1 - )]dz h h L0 3

= abr[ =

[

3 1 1 1 1 a2 (h4 - h4 + h4 - h4) + ( h4 - h4 )] 2 4 h 3 4 3h3

1 abhr(a2 + h2) 12

m = rV =

1 abhr 2

Thus, Iy =

m 2 (a + h2) 6

Ans.

Ans: Iy = 799

m 2 ( a + h2 ) 6

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17–10.

The pendulum consists of a 4-kg circular plate and a 2-kg slender rod. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O.

O

2m

Solution Using the parallel axis theorem by referring to Fig. a, IO = Σ ( IG + md 2 ) = c

1m

1 1 (2) ( 22 ) + 2 ( 12 ) d + c (4) ( 0.52 ) + 4 ( 2.52 ) d 12 2

= 28.17 kg # m2 Thus, the radius of gyration is kO =

28.17 IO = = 2.167 m = 2.17 m Am A4 + 2

Ans.

Ans: kO = 2.17 m 800

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17–11.

The assembly is made of the slender rods that have a mass per unit length of 3 kg>m. Determine the mass moment of inertia of the assembly about an axis perpendicular to the page and passing through point O.

O 0.4 m 0.8 m

0.4 m

Solution Using the parallel axis theorem by referring to Fig. a, IO = Σ ( IG + md 2 ) = e

1 33(1.2)4 ( 1.22 ) + 33(1.2)4 ( 0.22 ) f 12 + e

1 33(0.4)4 ( 0.42 ) + 33(0.4)4 ( 0.82 ) f 12

= 1.36 kg # m2

Ans.

Ans: IO = 1.36 kg # m2 801

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*17–12.

Determine the moment of inertia of the solid steel assembly about the x axis. Steel has a specific weight of gst = 490 lb>ft3.

0.25 ft 0.5 ft 2 ft

3 ft

SOLUTION Ix =

1 3 3 m (0.5)2 + m (0.5)2 m (0.25)2 2 1 10 2 10 3

1 3 1 3 1 490 a b p(0.5)2 (4)(0.5)2 a bp(0.25)2(2)(0.25)2 d a b = c p(0.5)2(3)(0.5)2 + 2 10 3 10 2 32.2 = 5.64 slug # ft2

Ans.

Ans: Ix = 5.64 slug # ft 2 802

x

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17–13.

The wheel consists of a thin ring having a mass of 10 kg and four spokes made from slender rods, each having a mass of 2 kg. Determine the wheel’s moment of inertia about an axis perpendicular to the page and passing through point A.

500 mm

SOLUTION A

IA = Io + md3 = c2c

1 (4)(1)2 d + 10(0.5)2 d + 18(0.5)2 12

= 7.67 kg # m2

Ans.

Ans: IA = 7.67 kg # m2 803

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17–14.

If the large ring, small ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A.

4 ft 1 ft O

SOLUTION Composite Parts: The wheel can be subdivided into the segments shown in Fig. a.

The spokes which have a length of (4 - 1) = 3 ft and a center of mass located at a 3 distance of a1 + b ft = 2.5 ft from point O can be grouped as segment (2).

2

A

Mass Moment of Inertia: First, we will compute the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O. IO = a

100 1 20 20 15 b (4 2) + 8c a b (32) + a b(2.52) d + a b(12) 32.2 12 32.2 32.2 32.2

= 84.94 slug # ft2

The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A can be found using the parallel-axis theorem 100 20 15 IA = IO + md 2, where m = + 8a b + = 8.5404 slug and d = 4 ft.

32.2 32.2 32.2 Thus, IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2

Ans.

Ans: IA = 222 slug # ft 2 804

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17–15.

Determine the moment of inertia about an axis perpendicular to the page and passing through the pin at O. The thin plate has a hole in its center. Its thickness is 50 mm, and the material has a density r = 50 kg>m3.

O

150 mm

SOLUTION IG =

1.40 m

1 1 C 50(1.4)(1.4)(0.05) D C (1.4)2 + (1.4)2 D - C 50(p)(0.15)2(0.05) D (0.15)2 12 2

1.40 m

= 1.5987 kg # m2 IO = IG + md2 m = 50(1.4)(1.4)(0.05) - 50(p)(0.15)2(0.05) = 4.7233 kg IO = 1.5987 + 4.7233(1.4 sin 45°)2 = 6.23 kg # m2

Ans.

Ans: IO = 6.23 kg # m2 805

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*17–16.

Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 20 kg>m2.

200 mm

O

200 mm

SOLUTION Composite Parts: The plate can be subdivided into two segments as shown in Fig. a.

Since segment (2) is a hole, it should be considered as a negative part. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated.

200 mm

Mass Moment of Inertia: The moment of inertia of segments (1) and (2) are computed as m1 = p(0.2 2)(20) = 0.8p kg and m2 = (0.2)(0.2)(20) = 0.8 kg. The moment of inertia of the plate about an axis perpendicular to the page and passing through point O for each segment can be determined using the parallel-axis theorem.

IO = ©IG + md2 1 1 = c (0.8p)(0.22) + 0.8p(0.22) d - c (0.8)(0.22 + 0.22) + 0.8(0.22) d 2 12 = 0.113 kg # m2

Ans.

Ans: IO = 0.113 kg # m2 806

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17–17.

Determine the location y of the center of mass G of the assembly and then calculate the moment of inertia about an axis perpendicular to the page and passing through G. The block has a mass of 3 kg and the semicylinder has a mass of 5 kg.

400 mm

300 mm G –y

Solution

200 mm O

Moment inertia of the semicylinder about its center of mass: (IG)cyc =

y =

4R 2 1 mR2 - m a b = 0.3199mR2 2 3p

Σ∼ ym Σm

=

c 0.2 -

4(0.2)

d (5) + 0.35(3) 3p = 0.2032 m = 0.203 m 5 + 3

IG = 0.3199(5)(0.2)2 + 5c 0.2032 - a0.2 -

4(0.2) 3p

2

bd +

Ans.

1 (3) ( 0.32 + 0.42 ) 12 + 3(0.35 - 0.2032)2

= 0.230 kg # m2

Ans.

Ans: IG = 0.230 kg # m2 807

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17–18.

Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point O. The block has a mass of 3 kg, and the semicylinder has a mass of 5 kg.

400 mm

300 mm G –y

Solution (IG)cyl =

200 mm O

2

1 4R mR2 - m a b = 0.3199 mR2 2 3p

IO = 0.3199(5)(0.2)2 + 5 a0.2 -

4(0.2) 3p

= 0.560 kg # m2

2

b +

1 (3) ( (0.3)2 + (0.4)2 ) + 3(0.350)2 12 Ans.

Also from the solution to Prob. 17–22, IO = IG + md 2 = 0.230 + 8(0.2032)2 = 0.560 kg # m2

Ans.

Ans: IO = 0.560 kg # m2 808

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17–19.

Determine the moment of inertia of the wheel about an axis which is perpendicular to the page and passes through the center of mass G. The material has a specific weight g = 90 lb>ft 3.

0.25 ft

1 ft

G

0.25 ft

2 ft

0.5 ft

Solution IG =

O 1 ft

1 90 1 90 c (p)(2)2(0.25) d (2)2 + c (p)(2.5)2(1) d (2.5)2 2 32.2 2 32.2 -

1 90 1 90 c b(p)(0.25)2(0.25) d (0.25)2 (p)(2)2(1) d (2)2 - 4c a 2 32.2 2 32.2

- 4 ca

90 b(p)(0.25)2(0.25) d (1)2 32.2

= 118.25 = 118 slug # ft 2

Ans.

Ans: IG = 118 slug # ft 2 809

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*17–20.

Determine the moment of inertia of the wheel about an axis which is perpendicular to the page and passes through point O. The material has a specific weight g = 90 lb>ft 3.

0.25 ft

1 ft

G

0.25 ft

2 ft

0.5 ft

Solution m =

O 1 ft

90 3p(2)2(0.25) + p 5 (2.5)2(1) - (2)2(1)6 - 4p(0.25)2(0.25)4 = 27.99 slug 32.2

From the solution to Prob. 17–18, IG = 118.25 slug # ft 2

IO = 118.25 + 27.99(2.5)2 = 293 slug # ft 2

Ans.

Ans: IO = 293 slug # ft 2 810

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17–21.

The pendulum consists of the 3-kg slender rod and the 5-kg thin plate. Determine the location y of the center of mass G of the pendulum; then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

O

2m

G

SOLUTION y =

y

0.5 m

1(3) + 2.25(5) © ym = = 1.781 m = 1.78 m ©m 3 + 5

Ans. 1m

IG = ©IG + md 2 =

1 1 (3)(2)2 + 3(1.781 - 1)2 + (5)(0.52 + 12) + 5(2.25 - 1.781)2 12 12

= 4.45 kg # m2

Ans.

Ans: y = 1.78 m IG = 4.45 kg # m2 811

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17–22.

20 mm

Determine the moment of inertia of the overhung crank about the x axis. The material is steel having a destiny of r = 7.85 Mg>m3.

30 mm 90 mm 50 mm x

180 mm

20 mm

SOLUTION

x¿

mc = 7.85(10 ) A (0.05)p(0.01) B = 0.1233 kg 3

30 mm

2

20 mm

mr = 7.85(103)((0.03)(0.180)(0.02)) = 0.8478 kg

50 mm

30 mm

1 Ix = 2 c (0.1233)(0.01)2 + (0.1233)(0.06)2 d 2 + c

1 (0.8478) A (0.03)2 + (0.180)2 B d 12

= 0.00325 kg # m2 = 3.25 g # m2

Ans.

Ans: Ix = 3.25 g # m2 812

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17–23.

20 mm

Determine the moment of inertia of the overhung crank about the x¿ axis. The material is steel having a destiny of r = 7.85 Mg>m3.

30 mm 90 mm 50 mm x

180 mm

20 mm

SOLUTION

x¿

mc = 7.85 A 103 B A (0.05)p(0.01)2 B = 0.1233 kg

30 mm

mp = 7.85 A 103 B A (0.03)(0.180)(0.02) B = 0.8478 kg

20 mm

50 mm

30 mm

1 1 Ix¿ = c (0.1233)(0.01)2 d + c (0.1233)(0.02)2 + (0.1233)(0.120)2 d 2 2 + c

1 (0.8478) A (0.03)2 + (0.180)2 B + (0.8478)(0.06)2 d 12

= 0.00719 kg # m2 = 7.19 g # m2

Ans.

Ans: Ix′ = 7.19 g # m2 813

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*17–24.

The door has a weight of 200 lb and a center of gravity at G. Determine how far the door moves in 2 s, starting from rest, if a man pushes on it at C with a horizontal force F = 30 lb.

Also, find the vertical reactions at the rollers A and B.

A

F

SOLUTION + ©F = m(a ) ; : x G x

6 ft

G

C

3 ft

200 )aG 30 = ( 32.2

6 ft

B

12 ft 5 ft

aG = 4.83 ft>s2 a+©MA = ©(Mk)A;

NB(12) - 200(6) + 30(9) = (

200 )(4.83)(7) 32.2 Ans.

NB = 95.0 lb + c ©Fy = m(aG)y ;

NA + 95.0 - 200 = 0 Ans.

NA = 105 lb + ) (:

s = s0 + v0t + s = 0 + 0 +

1 2 a t 2 G

1 (4.83)(2)2 = 9.66 ft 2

Ans.

Ans: NB = 95.0 lb NA = 105 lb s = 9.66 ft 814

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17–25.

The door has a weight of 200 lb and a center of gravity at G. Determine the constant force F that must be applied to the door to push it open 12 ft to the right in 5 s, starting from rest. Also, find the vertical reactions at the rollers A and B.

A

F

SOLUTION

12 = 0 + 0 +

6 ft

G

C

3 ft

+ )s = s + v t + 1 a t2 (: 0 0 G 2

6 ft

B

12 ft 5 ft

1 aG(5)2 2

ac = 0.960 ft>s2 + ©F = m(a ) ; : x G x

F =

200 (0.960) 32.2 Ans.

F = 5.9627 lb = 5.96 lb a + ©MA = ©(Mk)A ;

NB(12) - 200(6) + 5.9627(9) =

200 (0.960)(7) 32.2 Ans.

NB = 99.0 lb + c ©Fy = m(aG)y ;

NA + 99.0 - 200 = 0 Ans.

NA = 101 lb

Ans: F = 5.96 lb NB = 99.0 lb NA = 101 lb 815

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17–26.

The jet aircraft has a total mass of 22 Mg and a center of mass at G. Initially at take-off the engines provide a thrust 2T = 4 kN and T¿ = 1.5 kN. Determine the acceleration of the plane and the normal reactions on the nose wheel and each of the two wing wheels located at B. Neglect the mass of the wheels and, due to low velocity, neglect any lift caused by the wings.

T¿ 2.5 m

G

2T 2.3 m

1.2 m B 3m

6m

A

SOLUTION + ©F = ma ; : x x + c ©Fy = 0 ;

1.5 + 4 = 22aG 2By + Ay - 22(9.81) = 0

a + ©MB = ©(MK)B ;

4(2.3) - 1.5(2.5) - 22(9.81)(3) + Ay (9) = - 22aG(1.2) Ay = 72.6 kN

Ans.

By = 71.6 kN

Ans.

aG = 0.250 m>s2

Ans.

Ans: Ay = 72.6 kN By = 71.6 kN aG = 0.250 m>s2 816

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17–27.

The sports car has a weight of 4500 lb and center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft/s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are ms = 0.5 and mk = 0.3, respectively. Neglect the mass of the wheels.

G 2.5 ft B

4 ft

2 ft

A

SOLUTION a + ©MA = ©(Mk)A ; - 2NB(6) + 4500(2) =

- 4500 a (2.5) 32.2 G

+ ©F = m(a ) ; : x G x

0.3(2NB) =

+ c ©Fy = m(aG)y ;

2NB + 2NA - 4500 = 0

4500 a 32.2 G

Solving, NA = 1393 lb

Ans.

NB = 857 lb

Ans.

aG = 3.68 ft>s2 + ) (:

v = v0 + act 10 = 0 + 3.68 t Ans.

t = 2.72 s

Ans: NA = 1393 lb NB = 857 lb t = 2.72 s 817

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*17–28.

The assembly has a mass of 8 Mg and is hoisted using the boom and pulley system. If the winch at B draws in the cable with an acceleration of 2 m>s2, determine the compressive force in the hydraulic cylinder needed to support the boom.

The boom has a mass of 2 Mg and mass center at G.

6m

2m

C

4m

Solution

B

sB + 2sL = l 1m

aB = - 2aL 2 = -2aL

G

60 A

D

2m

aL = - 1 m>s2 Assembly: + c ΣFy = may; 2T - 8 ( 103 ) (9.81) = 8 ( 103 ) (1) T = 43.24 kN Boom: FCD(2) - 2 ( 103 ) (9.81)(6 cos 60°) - 2(43.24) ( 103 ) (12 cos 60°) = 0 a+ ΣMA = 0; Ans.

FCD = 289 kN

Ans: FCD = 289 kN 818

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17–29.

The assembly has a mass of 4 Mg and is hoisted using the winch at B. Determine the greatest acceleration of the assembly so that the compressive force in the hydraulic cylinder supporting the boom does not exceed 180 kN. What is the tension in the supporting cable? The boom has a mass of 2 Mg and mass center at G.

6m

2m

B

Boom: a + ΣMA = 0;

C

4m

Solution 180 ( 103 ) (2) - 2 ( 103 ) (9.81)(6 cos 60°) - 2T(12 cos 60°) = 0

1m

G

60 A

D

Ans.

T = 25 095 N = 25.1 kN

2m

Assembly: + c ΣFy = may ; 2(25 095) - 4 ( 103 ) (9.81) = 4 ( 103 ) a a = 2.74 m>s2

Ans.

Ans: a = 2.74 m>s2 T = 25.1 kN 819

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17–30.

The uniform girder AB has a mass of 8 Mg. Determine the internal axial, shear, and bending-moment loadings at the center of the girder if a crane gives it an upward acceleration of 3 m>s2.

3 m/s2 C

Solution Girder: + c ΣFy = may ; 2T sin 60° - 8000(9.81) = 8000(3)

A

T = 59 166.86 N

60

4m

60

B

Segment: + ΣFx = max ; 59 166.86 cos 60° - N = 0 S Ans.

N = 29.6 kN + c ΣFy = may ; 59 166.86 sin 60° - 4000(9.81) + V = 4000(3)

Ans.

V = 0

a+ ΣMC = Σ(Mk)C ; M + 4000(9.81)(1) - 59 166.86 sin 60°(2) = - 4000(3)(1) M = 51.2 kN # m

Ans.

Ans: N = 29.6 kN V = 0 M = 51.2 kN # m 820

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17–31. z

A car having a weight of 4000 lb begins to skid and turn with the brakes applied to all four wheels. If the coefficient of kinetic friction between the wheels and the road is mk = 0.8, determine the maximum critical height h of the center of gravity G such that the car does not overturn. Tipping will begin to occur after the car rotates 90° from its original direction of motion and, as shown in the figure, undergoes translation while skidding. Hint: Draw a free-body diagram of the car viewed from the front. When tipping occurs, the normal reactions of the wheels on the right side (or passenger side) are zero.

2.5 ft x h 2.5 ft

G

y

Solution NA represents the reaction for both the front and rear wheels on the left side. + ΣF = m(a ) ; 0.8N = 4000 a d A x G x 32.2 G + c ΣFy = m(aG)y ; NA - 4000 = 0 a + ΣMA = Σ(Mk)A; 4000(2.5) =

4000 (a )(h) 32.2 G

Solving, NA = 4000 lb aG = 25.76 ft>s2 Ans.

h = 3.12 ft

Ans: h = 3.12 ft 821

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*17–32.

A force of P = 300 N is applied to the 60-kg cart. Determine the reactions at both the wheels at A and both the wheels at B. Also, what is the acceleration of the cart? The mass center of the cart is at G.

P 30 G 0.4 m 0.3 m B

A

Solution

0.3 m

Equations of Motions. Referring to the FBD of the cart, Fig. a,

0.2 m

0.08 m

+ ΣF = m(a ) ; 300 cos 30° = 60a d x G x a = 4.3301 m>s2 = 4.33 m>s2 d

Ans.

+c ΣFy = m(aG)y ; NA + NB + 300 sin 30° - 60(9.81) = 60(0)

(1)

a+ΣMG = 0; NB(0.2) - NA(0.3) + 300 cos 30°(0.1) - 300 sin 30°(0.38) = 0

(2)

Solving Eqs. (1) and (2), NA = 113.40 N = 113 N

Ans.

NB = 325.20 N = 325 N

Ans.

Ans: a = 4.33 m>s2 d NA = 113 N NB = 325 N 822

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17–33.

Determine the largest force P that can be applied to the 60-kg cart, without causing one of the wheel reactions, either at A or at B, to be zero. Also, what is the acceleration of the cart? The mass center of the cart is at G.

P 30 G 0.4 m 0.3 m B

A

Solution Equations of Motions. Since (0.38 m) tan 30° = 0.22 m 7 0.1 m, the line of action of P passes below G. Therefore, P tends to rotate the cart clockwise. The wheels at A will leave the ground before those at B. Then, it is required that NA = 0. Referring, to the FBD of the cart, Fig. a + c ΣFy = m(aG)y; NB + P sin 30° - 60(9.81) = 60(0)

(1)

a+ ΣMG = 0; P cos 30°(0.1) - P sin 30°(0.38) + NB(0.2) = 0

(2)

0.3 m

0.2 m

0.08 m

Solving Eqs. (1) and (2) Ans.

P = 578.77 N = 579 N NB = 299.22 N

Ans: P = 579 N 823

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17–34.

The trailer with its load has a mass of 150 kg and a center of mass at G. If it is subjected to a horizontal force of P = 600 N, determine the trailer’s acceleration and the normal force on the pair of wheels at A and at B. The wheels are free to roll and have negligible mass.

G P

1.25 m 0.25 m

SOLUTION

0.75 m

Equations of Motion: Writing the force equation of motion along the x axis, + ©F = m(a ) ; : x G x

600 = 150a

a = 4 m>s2 :

B

0.25 m

A

600 N

0.5 m

1.25 m

Ans.

Using this result to write the moment equation about point A, a + ©MA = (Mk)A ;

150(9.81)(1.25) - 600(0.5) - NB(2) = -150(4)(1.25) Ans.

NB = 1144.69 N = 1.14 kN Using this result to write the force equation of motion along the y axis, + c ©Fy = m(a G)y ;

NA + 1144.69 - 150(9.81) = 150(0) Ans.

NA = 326.81 N = 327 N

Ans: a = 4 m>s2 S NB = 1.14 kN NA = 327 N 824

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17–35.

The desk has a weight of 75 lb and a center of gravity at G. Determine its initial acceleration if a man pushes on it with a force F = 60 lb. The coefficient of kinetic friction at A and B is mk = 0.2.

F 1 ft

30 G

2 ft A

Solution

+ ΣFx = max; 60 cos 30° - 0.2NA - 0.2NB = 75 aG S 32.2

B 2 ft

2 ft

+ c ΣFy = may; NA + NB - 75 - 60 sin 30° = 0 a+ΣMG = 0; 60 sin 30°(2) - 60 cos 30°(1) - NA(2) + NB(2) - 0.2NA(2) - 0.2NB(2) = 0 Solving, aG = 13.3 ft>s2

Ans.

NA = 44.0 lb NB = 61.0 lb

Ans: aG = 13.3 ft>s2 825

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*17–36.

The desk has a weight of 75 lb and a center of gravity at G. Determine the initial acceleration of a desk when the man applies enough force F to overcome the static friction at A and B. Also, find the vertical reactions on each of the two legs at A and at B. The coefficients of static and kinetic friction at A and B are ms = 0.5 and mk = 0.2, respectively.

F 1 ft

30 G

2 ft A

Solution

B 2 ft

Force required to start desk moving;

2 ft

+ ΣFx = 0; F cos 30° - 0.5NA - 0.5NB = 0 S + c ΣFy = 0; NA + NB - F sin 30° - 75 = 0 Solving for F by eliminating NA + NB, F = 60.874 lb Desk starts to slide. + ΣFx = m(aG)x; 60.874 cos 30° - 0.2NA - 0.2NB = 75 aG S 32.2 + c ΣFy = m(aG)y; NA + NB - 60.874 sin 30° - 75 = 0 Solving for aG by eliminating NA + NB, aG = 13.58 = 13.6 ft>s2

Ans.

a+ ΣMA = Σ(MK)A; NB(4) - 75(2) - 60.874 cos 30°(3) =

-75 (13.58)(2) 32.2

NB = 61.2 lb So that NA = 44.2 lb For each leg, ′ NA =

44.2 = 22.1 lb 2

Ans.

NB′ =

61.2 = 30.6 lb 2

Ans.

Ans: aG = 13.6 ft>s2 ′ NA = 22.1 lb NB′ = 30.6 lb 826

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17–37.

The 150-kg uniform crate rests on the 10-kg cart. Determine the maximum force P that can be applied to the handle without causing the crate to tip on the cart. Slipping does not occur.

0.5 m

P 1m

Solution Equation of Motion. Tipping will occur about edge A. Referring to the FBD and kinetic diagram of the crate, Fig. a, a+ ΣMA = Σ(MK)A; 150(9.81)(0.25) = (150a)(0.5) a = 4.905 m>s2 Using the result of a and refer to the FBD of the crate and cart, Fig. b, + ΣF = m(a ) P = (150 + 10)(4.905) = 784.8 N = 785 N d x G x

Ans.

Ans: P = 785 N 827

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17–38.

The 150-kg uniform crate rests on the 10-kg cart. Determine the maximum force P that can be applied to the handle without causing the crate to slip or tip on the cart. The coefficient of static friction between the crate and cart is ms = 0.2.

0.5 m

P 1m

Solution Equation of Motion. Assuming that the crate slips before it tips, then Ff = ms N = 0.2 N. Referring to the FBD and kinetic diagram of the crate, Fig. a + c ΣFy = may; N - 150 (9.81) = 150 (0) N = 1471.5 N + ΣF = m(a ) ; 0.2(1471.5) = 150 a a = 1.962 m>s2 d x G x a+ΣMA = (Mk)A; 150(9.81)(x) = 150(1.962)(0.5) x = 0.1 m Since x = 0.1 m 6 0.25 m, the crate indeed slips before it tips. Using the result of a and refer to the FBD of the crate and cart, Fig. b, + ΣFx = m(aG)x; P = (150 + 10)(1.962) = 313.92 N = 314 N d

Ans.

Ans: P = 314 N 828

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17–39.

The bar has a weight per length w and is supported by the smooth collar. If it is released from rest, determine the internal normal force, shear force, and bending moment in the bar as a function of x.

30

x

Solution Entire bar: ΣFx′ = m(aG)x′; wl cos 30° =

wl (a ) g G

aG = g cos 30° Segment: + ΣF = m(a ) ; N = (wx cos 30°) sin 30° = 0.433wx d x G x

Ans.

+ T ΣFy = m(aG)y; wx - V = wx cos 30°(cos 30°) Ans.

V = 0.25wx x x a+ ΣMS = Σ(Mk)S; wxa b - M = wx cos 30°(cos 30°)a b 2 2 M = 0.125wx2

Ans.

Ans: N = 0.433wx V = 0.25wx M = 0.125wx2 829

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*17–40.

The smooth 180-lb pipe has a length of 20 ft and a negligible diameter. It is carried on a truck as shown. Determine the maximum acceleration which the truck can have without causing the normal reaction at A to be zero. Also determine the horizontal and vertical components of force which the truck exerts on the pipe at B.

20 ft

A 5 ft

B

Solution

12 ft

+ ΣF = ma ; B = 180 a S x x x 32.2 T + c ΣFy = 0; By - 180 = 0 c+ΣMB = Σ(Mk)B; 180(10)a Solving,

180 12 5 aT (10)a b b = 13 32.2 13 Ans.

Bx = 432 lb

Ans.

By = 180 lb 2

Ans.

aT = 77.3 ft>s

Ans: Bx = 432 lb By = 180 lb aT = 77.3 ft>s2 830

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17–41.

The smooth 180-lb pipe has a length of 20 ft and a negligible diameter. It is carried on a truck as shown. If the truck accelerates at a = 5 ft>s2, determine the normal reaction at A and the horizontal and vertical components of force which the truck exerts on the pipe at B.

20 ft

A 5 ft

B

Solution

12 ft

+ ΣF = ma ; B - N a 5 b = 180 (5) S x A x x 13 32.2 + c ΣFy = 0 ; By - 180 + NAa

12 b = 0 13

a+ΣMB = Σ(Mk)B; - 180(10)a Solving,

12 180 5 (5)(10)a b b + NA(13) = 13 32.2 13

Bx = 73.9 lb

Ans.

By = 69.7 lb

Ans.

NA = 120 lb

Ans.

Ans: Bx = 73.9 lb By = 69.7 lb NA = 120 lb 831

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17–42.

The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration? The coefficient of static friction between the crate and the cart is ms = 0.5.

0.6 m F

1m

SOLUTION 15

Equations of Motion: Assume that the crate slips, then Ff = ms N = 0.5N. a + ©MA = ©(Mk)A ;

50(9.81) cos 15°(x) - 50(9.81) sin 15°(0.5) = 50a cos 15°(0.5) + 50a sin 15°(x)

(1)

+Q©Fy¿ = m(aG)y¿ ;

N - 50(9.81) cos 15° = -50a sin 15°

(2)

R + ©Fx¿ = m(aG)x¿ ;

50(9.81) sin 15° - 0.5N = -50a cos 15°

(3)

Solving Eqs. (1), (2), and (3) yields N = 447.81 N

x = 0.250 m

a = 2.01 m>s2

Ans.

Since x 6 0.3 m , then crate will not tip. Thus, the crate slips.

Ans.

Ans: a = 2.01 m>s2 The crate slips.

832

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17–43.

1 ft

Determine the acceleration of the 150-lb cabinet and the normal reaction under the legs A and B if P = 35 lb. The coefficients of static and kinetic friction between the cabinet and the plane are ms = 0.2 and mk = 0.15, respectively. The cabinet’s center of gravity is located at G. P

G 4 ft

SOLUTION Equations of Equilibrium: The free-body diagram of the cabinet under the static condition is shown in Fig. a, where P is the unknown minimum force needed to move the cabinet. We will assume that the cabinet slides before it tips. Then, FA = msNA = 0.2NA and FB = msNB = 0.2NB. + ©F = 0; : x

P - 0.2NA - 0.2NB = 0

(1)

+ c ©Fy = 0;

NA + NB - 150 = 0

(2)

+ ©MA = 0;

NB(2) - 150(1) - P(4) = 0

(3)

1 ft

3.5 ft

A

B

Solving Eqs. (1), (2), and (3) yields P = 30 lb

NA = 15 lb

NB = 135 lb

Since P 6 35 lb and NA is positive, the cabinet will slide.

Equations of Motion: Since the cabinet is in motion, FA = mkNA = 0.15NA and FB = mkNB = 0.15NB. Referring to the free-body diagram of the cabinet shown in Fig. b, + ©F = m(a ) ; : x G x + ©F = m(a ) ; : x G x + ©MG = 0;

35 - 0.15NA - 0.15NB = a

150 ba 32.2

(4) (5)

NA + NB - 150 = 0

NB(1) - 0.15NB(3.5) - 0.15NA(3.5) - NA(1) - 35(0.5) = 0

(6)

Solving Eqs. (4), (5), and (6) yields a = 2.68 ft>s2 NA = 26.9 lb

Ans. Ans.

NB = 123 lb

Ans: a = 2.68 ft>s2 NA = 26.9 lb NB = 123 lb 833