A construction worker attempts to lift a uniform beam off the floor and raise it to a vertical position. The beam is 2.50m long and weighs 500N. At a certain instant the worker holds the beam momentarily at rest with one end at distance d=1.50m above the floor by exerting a force P on the beam, perpendicular to the beam. What is the magnitude P?
It depends on where he is holding the beam, you did not state that.
He's pushing on it at the end that's not on the floor.
T= FpL- Wb(l/2)sin(53)
Fp = {Wb(l/2)sin(53)}/l
put the value,
Fp = 200 N
To find the magnitude of the force P exerted by the worker to hold the beam at rest, we can use the principle of moments.
The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.
In this case, we can take moments about the end of the beam resting on the floor. Let's call this point O. Since the beam is at rest, the sum of the moments about point O must be zero.
The weight of the beam acts downwards from the center of mass, which is at the midpoint of the beam. Since the beam is uniform, the weight can be considered as acting at this midpoint.
The distance of the midpoint from point O is 2.50m / 2 = 1.25m. The weight (500N) acts perpendicular to the beam.
To hold the beam at rest, the worker exerts a force P at a distance d = 1.50m above the floor.
Now, let's apply the principle of moments:
Sum of clockwise moments = Sum of anticlockwise moments
Clockwise moment due to weight = 500N × 1.25m = 625Nm (since the weight acts at a distance of 1.25m from point O in the opposite direction to the anticlockwise rotation)
Anticlockwise moment due to the worker's force P = P × 1.50m
Therefore, 625Nm = P × 1.50m
To find P, we can rearrange the equation:
P = 625Nm / 1.50m
P = 416.67N
Therefore, the magnitude of the force P exerted by the worker to hold the beam at rest is approximately 416.67N.