A uniform bar of length 3.5 m and mass 4.8 kg is attached to a wall through a hi

Question

A uniform bar of length 3.5 m and mass 4.8 kg is attached to a wall through a hinge mechanism which allows it to rotate freely. The other end of the bar is supported by a rope of length 6.6 m which is also connected to the wall as shown above. What is the tension in the rope in N?

M A uniform bar of length 3.5 m and mass 4.8 kg is attached to a wall through a hinge mechanism which allows it to rotate freely. The other end of the bar is supported by a rope of length 6.6 m which is also connected to the wall as shown above. What is the tension in the rope in N?

Explanation / Answer

Given that system is in equilibrium ,hence net torque acting on the system is zero

Tnet = 0

Torque due to weight of rod + tension = 0

[(-L/2)*W_rod] + (T*L*sin(theta)) = 0

(L/2)*W_rod = T*L*sin(theta)

L cancels

W_rod = 4.8*9.8 N

T is tension in rope

from the figure cos(theta) = 3.5/6.6

s

sin(theta) = sqrt(1-cos^2(theta)) = sqrt(1-(3.5/6.6)^2) = 0.85

then
(1/2)*W_rod = T*sin(theta)

Tension T = W_rod/(2*sin(theta)) = (4.8*9.8)/(2*0.85) = 27.67 N

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