A construction worker attempts to lift a uniform beam off the floor and raise it

Question

A construction worker attempts to lift a uniform beam off the floor and raise it to a vertical position. The beam is 2.47 m long and weighs 437 N. At a certain instant the worker holds the beam momentarily at rest with one end a distance d = 1.54 m above the floor, as shown in the figure, by exerting a force Upper P Overscript right-arrow EndScripts on the beam. (a) What is the magnitude of Upper P Overscript right-arrow EndScripts? (b) What is the magnitude of the (net) force of the floor on the beam? (c) What is the minimum value the coefficient of static friction between beam and floor can have in order for the beam not to slip at this instant?

Explanation / Answer

given that

length of beam    l = 2.47 m

weight = 437 N

d = 1.54 m

part(a)

the beam is at rest So

net torque = 0.

Net torque = (437N)*(1.54m)(cos(theta)) + (P)*(2.47m)(sin90)

from the given condition    sin(theta) = 1.54/2.47

theta = 38.53 degree

So
P = (437*1.54*cos(38.53)) / 2.47

P = 213.14 N

part(B)

the beam at rest so som of verticle forces is zero.

P*cos(38.53) + N = 437

N = 270.27 N

part(c)

the beam at the rest so sum of horizontal forces is zero .

P*sin(38.53) + friction force = 0

132.77 + u*N = 0

u*N = -132.77

negative sign indicates the direction opposite to the horizontal component of force F .

u = -132.77 /270.27

u = 0.49

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