A uniform horizontal bar of mass 1.78 kg and length 1.74 m is held by a friction

Question

A uniform horizontal bar of mass 1.78 kg and length 1.74 m is held by a frictionless pin at a wall. The opposite end of the strut is supported by a cord with tension T at an angle ?. A block of mass 3.56 kg is hung from the bar at a distance of 3/4 L from the pin.
A) Find the magnitude of the horizontal component of the force of the wall acting on the bar if the string makes an angle of 38.4? with the horizontal.
B) What is the direction of the horizontal component of the force of the wall acting on the bar? right or left

Explanation / Answer

A) equating torques about the pin,
Fsin38.4*l = 1.78*9.8*(l/2) + 3.56*9.8* (3/4)l
so, F = 56.2N

so, horizontal component of the force on the wall = 56.2*cos38.4 = 44N

B)acting left (the diagram is not provided so as i hav taken the pin to be at the left end, the answer is towards left...if the pin is on the left,then the answer is towards right )

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