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## Homework Statement

A construction worker attempts to lift a uniform beam off the floor and raise it to a vertical position. The beam is 2.26 m long and weighs 397 N. At a certain instant the worker holds the beam momentarily at rest with one end a distance

*d*= 1.21 m above the floor, as shown in the figure, by exerting a force on the beam.

**(a)**What is the magnitude of ?

**(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?

## Homework Equations

Just equilibrium conditions. Net force and net torque of the system are both zero.

## The Attempt at a Solution

So I figured out a)168N and c)0.35 but can't find part b

I tried equilibrium conditions for the vertical forces.

W = 397N

F = 168N

N = normal force

theta = 32.37 degrees

W = Fcos(theta) + N

Solving for normal force I got

397N - 141N = 255N

This answer is wrong. What's confusing me is if I use this value for part C to find the coefficient of static friction it works just fine. The answer isn't zero, and even if it was, that would be wrong because the floor pushes back up on the beam due to molecular compression. I don't get why it isn't 255N? Did I miss something here?

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