A person with weight 800 N stands d = 3.00 m away from the wall on a f = 6.00 m

Question

A person with weight 800 N stands d = 3.00 m away from the wall on a f = 6.00 m beam, as shown in this figure. The weight of the beam is 2,000 N. Define upward as the positive y direction and to the right as the positive x direction. (a) Find the tension in the wire. (Enter the magnitude only.) Find the net torque and the components of net torque acting, and apply the condition for the equilibrium to find the tension in the cable. N (b) Find the horizontal component of the hinge force. (Indicate the direction with the sign of your answer.) N (c) Find the vertical component. (Indicate the direction with the sign of your answer.) N

Explanation / Answer

Given w1 = 800 N

w2 = 2000 N

d = 3m   and   l = 6 m

torque T = r * F * sin(theta)

the net torque about the hinge is

(d * w1 * sin(270)) + (l/2 * w2 * sin(270)) + (l * T * sin(30)) = 0

(3 * 800 * sin(270)) + (3 * 2000 * sin(270)) + (6 * T * sin(30)) = 0

tension T = 2800 N

b) horizontal component is Tx = T * cos(30)

Tx = 2800 * cos(30)

Tx = 2424.87 N-------------------------------(towards left)

c) vertical component is Ty = T * sin(30)

Ty = 2800 * 0.5

Ty = 1400 N-----------------------(towards down)

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