A uniform beam of length 8.00m, and weight 200N is Attached to a wall by as pin connection. Its far end is supported by cable that makes an angle of 53 degrees with the horizontal. Find the tension in the cable and the force exerted by the wall on the beam.
Answers
Answer:
Explanation:
Given A uniform horizontal beam with a length of 8.00 m and a weight of 200 n is attached to a wall by a pin connection ( allows rotation). Its far end is supported by a cable that makes an angle of 53 degrees with the horizontal. If a 600 n person stands on the beam at 2.00 m from the wall , find the tension in the cable
Now the force acting on the beam are the gravitational force of earth and weight attached to a wall is 200 N which acts at centre. Also force of person standing is 600 N and force exerted is r and tension is T , also torque due to r is zero because it acts at the pivot point. So r = 0
Torque due to tension in the rope is sin 53 x 8.00 x T.
Torque due to force is 2.0 m x 600 N
The weight of beam acts in the centre since the beam is uniform.
So we have
sin 53 x 8.00 x T - 2 m x 600 N - 4 x 200 N = 0
8 .00 x 0.8 x T = 1200 + 800
0.64 T = 2000 N-m
T = 2000 / 0.64
T = 312.5 N
Answer:
Explanation:
A uniform horizontal beam with a length of 8.00 m and a weight of 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle 53.0owith the horizontal. If a 600-N person stands 2.00 m from the wall, find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam