a box A box (mass 5.6 kg) sits 3 meters from the right hand side ot a 10 meter l

Question

a box

A box (mass 5.6 kg) sits 3 meters from the right hand side ot a 10 meter long beam (mass 20 kg). The beam is supported on the left by a rope that is 0.5 meters from the left end of the beam. On the right edge, Greg supports the beam on his head. Draw an extended free body diagram tor the beam. Determine the magnitude of the normal force by Greg's head on the beam. Find the tension in the rope. The rope will break if more than 150N of force is applied to it. If the box is slid to the left, the tension in the rope increases. How far can we slide the box (from its starting position) before the rope breaks?

Explanation / Answer

a) there are total 4 forces on beam:

1.due to rope vertically upwards - T at rope point

2. due to weight of beam - mg - at the centre of beam (midpoint)vertically downwards

3.due to box - vertical downwards at the centre of box

4. due to head of gregg - normal force at the head point.

b)

as beams is in equilibrium so net torque will be zero.

balancing torque(r x F) about rope point

((5-0.5) x 20g) + ((10-3-0.5) x 5.6g) - ( (10 -0.5) x N) = 0

N = 130.53 N

c)

for equilibrium
Fnet = 0

so T - mg - Mg + N = 0

T - (20 x 9.81) - (5.6 x 9.81) + 130.53 = 0

T = 120.61 N .....>>Ans

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