Asked by TucsonGuy
An A-frame made up of two uniform 150N, 3 m long beams pinned at the apex and resting on a frictionless floor. The base of each beam is separated by 3.5 m. A 500 N weight hangs from the apex. A tie rope connecting the two beams is parallel to the floor and is attached to each beam 0.5 m above the floor.
What is the tension in the rope. Please give details, not a general solution.
What is the tension in the rope. Please give details, not a general solution.
Answers
Answered by
ajayb
Consider 'Free body diagram' of one of the beams (say left one) i.e. isolate it and see the forces acting on it.
Let's say end B is on the ground and end A is hinged to the other beam of the frame. Which are the forces acting on it?
1.Normal reaction N (vertically upwards)
2.Its weight(W=150N) acting@center (vertically downwards)
3.Weight 250N @end A(vertically downwards)(500N is equally shared by the two beams)
4.Tension T of the rope( horizontally towards right and 0.5m above ground)
Since the frame is in equlibrium:
N=150+250 = 400N
Take moment of these forces about point A.Torques produced in either direction should balance for equilibrium.
So
N*1.75 = 150*1.5*cos(theta)+ T*{3*sin (theta) -0.5}
here, T is tention in the rope and theta is the angle made by the beam with the horizontal and cos(theta) = 1.75/3 = 0.583 so theta = 54.3 deg
Plug in the value of N in the above equation and solve for T.
T = 294N
Let's say end B is on the ground and end A is hinged to the other beam of the frame. Which are the forces acting on it?
1.Normal reaction N (vertically upwards)
2.Its weight(W=150N) acting@center (vertically downwards)
3.Weight 250N @end A(vertically downwards)(500N is equally shared by the two beams)
4.Tension T of the rope( horizontally towards right and 0.5m above ground)
Since the frame is in equlibrium:
N=150+250 = 400N
Take moment of these forces about point A.Torques produced in either direction should balance for equilibrium.
So
N*1.75 = 150*1.5*cos(theta)+ T*{3*sin (theta) -0.5}
here, T is tention in the rope and theta is the angle made by the beam with the horizontal and cos(theta) = 1.75/3 = 0.583 so theta = 54.3 deg
Plug in the value of N in the above equation and solve for T.
T = 294N