210
Basic Engineering Mathematics
The magnitude of the resultant of vector addition = H 2 + V 2 = 9.8282 + 2.8282 √ = 104.59 = 10.23 N
4. Three forces of 2 N, 3 N and 4 N act as shown in Fig. 27.10. Calculate the magnitude of the resultant force and its direction relative to the 2 N force. 4N
The direction of the resultant of vector addition V 2.828 = tan−1 = 16.05◦ = tan−1 H 9.828 Thus, the resultant of the two forces is a single vector of 10.23 N at 16.05◦ to the 7 N vector. 60°
Problem 5. Calculate the resultant velocity of the three velocities given in Problem 2.
60°
With reference to Fig. 27.5: Horizontal component of the velocity, H = 10 cos 20◦ + 15 cos 90◦ + 7 cos 190◦ = 9.397 + 0 + (−6.894) = 2.503 m/s Vertical component of the velocity, V = 10 sin 20◦ + 15 sin 90◦ + 7 sin 190◦ = 3.420 + 15 + (−1.216) = 17.204 m/s Magnitude of the resultant of vector addition = H 2 + V 2 = 2.5032 + 17.2042 √ = 302.24 = 17.39 m/s Direction of the resultant of vector addition V 17.204 = tan−1 = tan−1 H 2.503 = tan−1 6.8734 = 81.72◦ Thus, the resultant of the three velocities is a single vector of 17.39 m/s at 81.72◦ to the horizontal
3N
2N
Fig. 27.10
5. A load of 5.89 N is lifted by two strings, making angles of 20◦ and 35◦ with the vertical. Calculate the tensions in the strings. [For a system such as this, the vectors representing the forces form a closed triangle when the system is in equilibrium]. 6. The acceleration of a body is due to four component, coplanar accelerations. These are 2 m/s2 due north, 3 m/s2 due east, 4 m/s2 to the south-west and 5 m/s2 to the south-east. Calculate the resultant acceleration and its direction. 7. A current phasor i1 is 5 A and horizontal.A second phasor i2 is 8 A and is at 50◦ to the horizontal. Determine the resultant of the two phasors, i1 + i2 , and the angle the resultant makes with current i1 . 8. An object is acted upon by two forces of magnitude 10 N and 8 N at an angle of 60◦ to each other. Determine the resultant force on the object. 9. A ship heads on a course due south at 12 km/h. It is moved off course by a current flowing in a north-easterly direction at 8 km/h. Determine the resultant velocity.
Now try the following exercise Exercise 98 Further problems on vectors (Answers on page 281) 1. Forces of 23 N and 41 N act at a point and are inclined at 90◦ to each other. Find, by drawing, the resultant force and its direction relative to the 41 N force. 2. Forces A, B and C are coplanar and act at a point. Force A is 12 kN at 90◦ , B is 5 kN at 180◦ and C is 13 kN at 293◦ . Determine graphically the resultant force. 3. Calculate the magnitude and direction of velocities of 3 m/s at 18◦ and 7 m/s at 115◦ when acting simultaneously on a point.
27.4 Vector subtraction In Fig. 27.11, a force vector F is represented by oa. The vector (−oa) can be obtained by drawing a vector from o in the opposite sense to oa but having the same magnitude, shown as ob in Fig. 27.11, i.e. ob = (−oa). For two vectors acting at a point, as shown in Fig. 27.12(a), the resultant of vector addition is os = oa + ob. Fig. 27.12(b) shows vectors ob + (−oa), that is, ob − oa and the vector equation is ob − oa = od. Comparing od in Fig. 27.12(b) with the broken line ab in Fig. 27.12(a) shows that the second diagonal of the ‘parallelogram’ method of vector addition gives the magnitude and direction of vector subtraction of oa from ob.