Question

1- What is torque? 2- What are the conditions for the equilibrium? 3- A uniform rod of length L-10 m, weight 490 N is used to support an object with weight Wi-343 N at a distance of x,-4 m and a second load of weight W2-196 N at a distance of x2-9 m. The rod is supported by a cable which is attached at x- 6 m and acting at an angle of 30 degrees with respect to the rod (a) What is the required tension in the cable? (b) Find the net horizontal force (magnitude and direction) that the pivot at the wall exerts in this loaded condition (c) Find the net vertical force (magnitude and direction) that the pivot at the wall exerts in this loaded condition. W, = m,8

Answer 1

Torque T = F vector middot r vector Net force = 0 Net torque = 0 a) Let tension be T, then balancing torque about point of point, T sin theta middot x_T = W_1 x_1 + W_b middot L/2 + W_2 x_2 rightwardsdoublearrow T middot sin 30 degree middot 6 = 343 times 4 + 490 times 10/2 + 196 times 9 rightwardsdoublearrow T middot 1/2 times 6 = 1392 + 2450 + 1764 rightwardsdoublearrow T = 5586/3 = 1862 N rightwardsdoublearrow T = 1862 N b) Horizontal force, F_H, Magnitude, F_H = T cos 30 degree = 1862 times squareroot 3/2 = 1612.54 N Direction in +ve x-direction. c) Let vertical form be F_V in upward direction then F_v + T sin 30 degree = W_1 + W_2 + W_b rightwardsdoublearrow F_v + 1862 times 1/2 = 343 + 196 + 490 rightwardsdoublearrow F_v + 931 = 1029 rightwardsdoublearrow F_v = 98 N And this force is in upward direction i.e, +ve y-axis.