Find the frictional force between a block and a wall

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A block weighing 22 N is pushed against a wall with a force of 60 N at a 30° angle, and the coefficients of static and kinetic friction are 0.55 and 0.38, respectively. The normal force is calculated as 60 cos(30), and the static friction force is determined to be less than or equal to the product of the static friction coefficient and the normal force. The discussion reveals that the static friction force is not simply equal to μ_s N, as it must be less than or equal to that value. The final conclusion indicates that the static friction force is -8 N, suggesting a misunderstanding in the initial calculations. Understanding the relationship between static friction and the normal force is crucial for accurate problem-solving in physics.
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1. Homework Statement

A block weighing 22 N is pushed against a vertical wall by a force of magnitude 60 N directed 30° above the horizontal. The block is intitally at rest. The coefficient of static friction between the wall and the block is 0.55 and the coefficient of kinetic friction between them is 0.38.What is the force of friction between the block and the wall? Express your answer as positive if the force of friction points up.

Homework Equations


f=m*a

The Attempt at a Solution


netforcey = 60sin(30)-mg+0.55*normalforce
netforcex = 60cos(30)-normalforce
since the block is not moving in x direction a = 0 so normalforce=60cos(30)
since looking for just friction and block initially at rest normalforce*staticfrictioncoefficient = friction force
since gravity point down friction force point up 0.55*normalfoce = 29 NEWTONSi didn't mean to post pic with -29 NEWTONS i meant 29 NEWTONS but either way still wrong
 

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Hint: ##\mu_s N## does not necessarily equal the static friction force. Why not?
(Here ##N## represents the normal force.)
 
TSny said:
Hint: ##\mu_s N## does not necessarily equal the static friction force. Why not?
(Here ##N## represents the normal force.)
because static friction just has to be less than or equal to μsN?
 
isukatphysics69 said:
because static friction just has to be less than or equal to μsN?
Yes. Good.
 
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TSny said:
Yes. Good.
OK got it the static friction is -8N thank you
 
:thumbup:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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