Construction Worker Lifting Uniform Beam

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SUMMARY

The discussion centers on a construction worker lifting a uniform beam weighing 397 N and measuring 2.26 m in length. The worker holds the beam at a height of 1.21 m while exerting a force of 168 N. The net force of the floor on the beam was initially miscalculated, with the correct approach requiring the consideration of both the normal force and the x-component of the applied force. The minimum coefficient of static friction necessary to prevent slipping was determined to be 0.35.

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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.
Transtutors001_d6995745-e53b-4939-96ea-1b4f3886a4bd.PNG

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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Fetch said:
Solving for normal force I got
397N - 141N = 255N
That's all correct, but (b) does not ask for the normal force.
 
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Yeah I misconstrued the (net) force part. I figured it out now. I needed the normal force (N) and the x-component of the push (F) to find the net force on the beam. My mistake. Thank you.
 

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