Finding the Horizontal Force on a Bracket Supported by a Single Screw

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Homework Help Overview

The problem involves a bracket supported by a single screw on a vertical wall, with a vertical force applied. Participants are tasked with finding the horizontal component of the force exerted by the screw on the bracket, while ignoring the weight of the bracket itself.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to analyze the problem using torque calculations about different rotational axes, questioning which approach is correct. Some participants inquire about the presence of a reaction force at point B, while others suggest summing forces in each direction to clarify the situation.

Discussion Status

Participants are exploring various interpretations of the forces involved and the application of Newton's Laws. There is a mix of attempts to calculate moments and discussions about horizontal forces, with no clear consensus reached yet.

Contextual Notes

Some participants express uncertainty regarding the existence of horizontal forces and the implications of their calculations, indicating a need for further clarification on the assumptions made in the problem setup.

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Homework Statement


One side of a plant shelf
is supported by a bracket
mounted on a vertical wall
by a single screw. Ignore
the weight of the bracket.
Find the horizontal
component of the force
that the screw exerts on
the bracket when an 80.0 N
vertical force is applied as
shown

image.jpg


Homework Equations

The Attempt at a Solution


If I choose A as rotational axis, then it should be calculated as

net torque = 0
(80N)(0.05m) = (x)(0.03m)

but if a choose B as the rotational axis, then it should be calculated as

net torque = 0
(80N)(0.05m) = (x)(0.06m)

So which one is correct?

Thanks!
 
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Is there a reaction @ B?
 
I don't think so. What does it mean?
 
Well, why don't you prove it and sum forces in each direction?
 
paisiello2 said:
Well, why don't you prove it and sum forces in each direction?

That's what I can think of.
(80N)(0.05m) = (n)(0.03m)
n = 133N

but this is incorrect and I don't know why

image.jpg
 
You're taking moments. I want you to sum forces in each direction: ΣFx = 0 and ΣFy=0.
 
paisiello2 said:
You're taking moments. I want you to sum forces in each direction: ΣFx = 0 and ΣFy=0.

Just Fx = 80 from the nail? I don't think there is a horizontal force.
image.jpg
 
Just apply Newton's Laws and you will see that there must be a horizontal force.
 

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