Question

A rod uniform and homogeneous AB of length L=4.00 m is attached to a wall as shown and secured via a massless rope BC of length l=5.00 ml=5.00 m to the wall (see Fig. 4). A mass M1=40.0 kg is attached to the rod at B. The rod itself has a mass M=2.00 kg. Assume that the tension in the string is T. Hint - set up the two force equations (along x and y axes) and the torque equation. The system is in equilibrium.

A) What is Hx?
B) What is Hy?
C) What is T?

(3 points) A rod uniform and homogeneous AB of length L-4.00 m is attached to a wall as shown and secured via a massless rope BC of length 15.00 to the wall (see Fig. 4). A mass Mi-40.0 kg is attached to the rod at B. The rod itself has a mass M-2.00 kg Assume that the tension in the string is T. Hint-set up the two torce equations (along z and y axes) and the torque equation. The system is in equilibrium. A) What is H? B) What is H? C) What is T? .00 ーし: 4.00 m FIG 4

Answer 1

Part A) - 535.73 N

Part B) - 9.798 N

Part C) 669.67 N

l = 5.00 m L = 4.00 m h = squareroot l^2 - L^2 h = 3m tan theta = h/L = 3/4 theta = 36.8698976 degree, M = 2.0 kg, M_1 = 40 kg force along horizontal direction H_x = -T cos theta = Tx Tx = Hx = -T times 4/5 = -4T/5 1 force along vertical direction Hg = T sin theta - Mg - M_1 g Hg = 3T/5 - 2 times 9.87 - 40 times 9.8 Hg = 3T/5 - 411.6 2 at equilibrium moment about A is zero T = sin theta times L - M_1 g times L - Mg times L/2 = 0 T times 3/5 times 4 = 40 times 9.8 times 4 + 2 times 9.8 times 2 T = 669.67 N part a from equation 1 H_x = -4/5 times 669.67 mu_x = -535.73 N \r\n part b from eqn 2 Hg = 3/5 times 669.67 - 411.6 Hg = -9.798 N