The drawing shows a uniform horizontal beam attached to a vertical wall by a fri
ID: 1289504 • Letter: T
Question
The drawing shows a uniform horizontal beam attached to a vertical wall by a frictionless hinge and supported from below at an angle 41 degrees by a brace that is attached to a pin. The beam has a weight of 342 N. Three additional forces keep the beam in equilibrium. The brace applies a force P to the right end of the beam that is directed upward at the angle theta, with respect to the horizontal. The hinge applies a force to the left end of the beam that has a horizontal component H and a vertical component V. Find the magnitudes of these three forces.
The drawing shows a uniform horizontal beam attached to a vertical wall by a frictionless hinge and supported from below at an angle 41 degrees by a brace that is attached to a pin. The beam has a weight of 342 N. Three additional forces keep the beam in equilibrium. The brace applies a force P to the right end of the beam that is directed upward at the angle theta, with respect to the horizontal. The hinge applies a force to the left end of the beam that has a horizontal component H and a vertical component V. Find the magnitudes of these three forces.Explanation / Answer
weight of the beam W =342 N
angle =41 deg
from figure :
horizontal force , Pcos?-F_H =0 ...... (1)
vertical force Psin?+F_V-W =0 ....... (2)
consider length of the beam is L and the lever arms for weight (W) and vertical force (F_V) are L/2 and L
respectively.here F_V and F_H are vertical and horizontal forces respectively.
the forces P and F_H creates torques relative to this axis,because their lines of action pass directly
through it.
there fore torques ,W(1/2)L-F_V(L) =0
1/2(W) =F_V
F_V =1/2(342 N)
F_V =171 N ..... (3)
substitute this value in equation (2) ,we get
P(sin41)+(171 N)-(342 N) =0
P =260.64 N
substitute theis value in equation (1) ,we get
(260.64N)cos41 -F_H =0
F_H =196.7 N
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