Question

# A bar of length L and uniform density is held horizontally against a wall by a...

A bar of length L and uniform density is held horizontally against a wall by a pivot at one end and a rope of tension T which pulls up and to the left making an angle \theta with the vertical wall. The rope is attached to the bar at a distance l = \frac{1}{3} L from the wall. A mass M is resting at the end of the bar.If the bar has total mass 2.6kg, the mass M=2.6kg, and the rope makes an angle of 22.0 ^{\circ} with the wall, what is the magnitude of the force exerted on the bar by the pivot | \vec{F_{pivot}} |? Answer in Newtons (N).

Given is:-

mass of the bar = 2.6kg

M = 2.6kg

Angle made by rope with the wall = 22 degrees.

where L = length of the bar

Since the bar is in equilibrium so the net torque about the pivot point must be zero.

Therefore the torque about pivot point is =

equating this to zero

by plugging all the values we get

which gives

Now,

The net vertical force at pivot =

Net vertical force at pivot = -63.7 N (negative sign shows that it is downwards)

Similarly the net horizontal force =

Net horizontal force at pivot = 46.326N

Hence the net force on pivot exerted by the bar is

by plugging all the values we get

we get

Therefore the required answer is

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