The drawing shows a uniform horizontal beam attached to a vertical wall by a fri

Question

The drawing shows a uniform horizontal beam attached to a vertical wall by a frictionless hinge and supported from below at an angle ? = 39o by a brace that is attached to a pin. The beam has a weight of 348 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 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.

I found that V =174 N and P =276.48 N. H=?????

Explanation / Answer

weight of the beam W =348 N angle =39 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(348 N) F_V =174 N ..... (3) ................................................................. substitute theis value in equation (2) ,we get P(sin39)+(174 N)-(348 N) =0 P =276.48 N .................................................................................... substitute theis value in equation (1) ,we get (276.48 N)cos39-F_H =0 F_H =214.86 N

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