The figure below depicts a uniform horizontal bar attached...

The figure below depicts a uniform horizontal bar attached to a vertical wall by a frictionless pivot and supported from below at an angle ? = 39° by a slanted brace that is attached to a pivot. The bar has a net weight of 340 N. Three more forces keep the beam in static equilibrium. The bracket applies a force P to the right end of the bar that is oriented upward at the angle ? relative to the horizontal. The pivot applies a force to the left end of the bar that has a horizontal component H and a vertical component V. Determine the magnitudes of these three forces.

Transcript

00:01 Hello students in this question we are given a figure and in this we are shown a horizontal bar attached to a vertical vault by a fissionist private and supported from below at an angle theta which is equal to 39 degree and by a slanted brace that is attached to the private and the bar has a net weight of 340 newton so weight of the bar is 340 newton and we have to find determine the three forces that is vh and p.

00:31 So let us see the diagram.

00:34 So this is the type of arrangement over here.

00:36 Now since the bar is in equilibrium, so we can write that summation of net force.

00:42 So net force will be equal to zero and net torque that is tau net will be equal to zero since the whole system is in equilibrium.

00:54 Now let the length of the bar be l.

00:56 Let the length of the length of bar b l over here now first let us consider the torque of the forces through the hinge will be zero as it passes through the hinge now so torque through hinge so it will be here so it will be p multiple the distance l and sine theta because here we have a vertical component and a horizontal component minus here the weight so w multiply by distance is l over 2 because it will be at center of mass so it will be l over 2 and this will be equal to 0 because the system is in equilibrium so from here we have p is equal to w over 2 sine theta so we have 340 over 2 sine 39 degree so here we have 270 .133 newton.

02:11 This is the value of p.

02:12 Now balancing the force along the horizontal so we can write h that is towards the left is equal to...

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