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
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.
1 answer