Coefficient of kinetic friction of a block sliding across the ceiling

[latos]

Homework Statement

A 4.00kg block is pushed along the ceiling with a constant applied force of 85.0N that acts at an angle of 55.0 degrees with the horizontal. The block accelerates to the right at 6.00 m/s^2. Determine the coefficient of kinetic friction between the block and ceiling.

Relevant Equations

F = ma
Ff = Uk*Fn

I first found the force of friction by setting the force in the x direction (horizontal component of force applied - friction) = ma. I then thought the normal force would be equal to vertical component of the force applied because of Newton's 3rd Law, which states that for every action, there is an opposite an equal reaction. I then plugged my friction and normal forces into the equation for kinetic friction and got .356, but apparently it is supposed to be .816. My teacher said you have to account for gravity in the normal force by subtracting the force of gravity from the vertical component. This made me confused because I thought the force applied would be directly transferred to the ceiling. Does this imply that 9.8m/s^2 * 4.00kg = 39.2N of force is lost when applying the force to the ceiling through the block instead of applying it directly to the ceiling?

 

[Aurelius120]

I then thought the normal force would be equal to vertical component of the force applied because of Newton's 3rd Law, which states that for every action, there is an opposite an equal reaction.

Why would Normal force be equal to vertical component of applied force?
For zero acceleration of block in vertical direction sum of $mg$, $N$ and the vertical component of the applied force should be zero.

 

[latos]

That is the mathematical reason, but I don't get why the force applied isn't physically being transferred directly to the ceiling. If pushed directly on the ceiling with 85N diagonally, I would get 85N towards me because of Newton's 3rd law. My question was why does the same not hold true if apply 85N to the block into the ceiling instead of directly into the ceiling? It seems like some force is being "lost" when I apply it through the block.

 

[Aurelius120]

That is the mathematical reason, but I don't get why the force applied isn't physically being transferred directly to the ceiling. If pushed directly on the ceiling with 85N diagonally, I would get 85N towards me because of Newton's 3rd law. My question was why does the same not hold true if apply 85N to the block into the ceiling instead of directly into the ceiling?

@latos You apply force on block. Block applies reaction force on you.

Block exerts action Normal on ceiling. Ceiling exerts reaction Normal on block.

Why will your action force on block be equal to normal reaction exerted by ceiling?

Since Normal force is repulsive, it can't pull block towards ceiling and since the block has weight, it must be balanced by these forces or it would lose contact with the ceiling and no friction will act on it.

It seems like some force is being "lost" when I apply it through the block.

I don't think force is conserved. Haven't heard of it so if it exists, I won't comment on it.

 

[SammyS]

Homework Statement: A 4.00kg block is pushed along the ceiling with a constant applied force of 85.0N that acts at an angle of 55.0 degrees with the horizontal. The block accelerates to the right at 6.00 m/s^2. Determine the coefficient of kinetic friction between the block and ceiling.
Relevant Equations: F = ma
Ff = Uk*Fn

View attachment 339658
I first found the force of friction by setting the force in the x direction (horizontal component of force applied - friction) = ma. I then thought the normal force would be equal to vertical component of the force applied because of Newton's 3rd Law, which states that for every action, there is an opposite an equal reaction. I then plugged my friction and normal forces into the equation for kinetic friction and got .356, but apparently it is supposed to be .816. My teacher said you have to account for gravity in the normal force by subtracting the force of gravity from the vertical component. This made me confused because I thought the force applied would be directly transferred to the ceiling. Does this imply that 9.8m/s^2 * 4.00kg = 39.2N of force is lost when applying the force to the ceiling through the block instead of applying it directly to the ceiling?

Hollo, @latos .


It is virtually impossible to read your workings in that posted image.

Your teacher is correct. Gravity is an important force in this problem?

What are all the forces being applied to the block ?

 

[Aurelius120]

@SammyS Here's the question

The working I can't read clearly either

 

[Lnewqban]

@SammyS Here's the question View attachment 339665
The working I can't read clearly either

Welcome!
Could you produce a free body diagram for the block alone?

My teacher said you have to account for gravity in the normal force by subtracting the force of gravity from the vertical component. This made me confused because I thought the force applied would be directly transferred to the ceiling. Does this imply that 9.8m/s^2 * 4.00kg = 39.2N of force is lost when applying the force to the ceiling through the block instead of applying it directly to the ceiling?

How can that applied force on the low surface of the block be measured?

 

[PeroK]

That is the mathematical reason, but I don't get why the force applied isn't physically being transferred directly to the ceiling. If pushed directly on the ceiling with 85N diagonally, I would get 85N towards me because of Newton's 3rd law. My question was why does the same not hold true if apply 85N to the block into the ceiling instead of directly into the ceiling? It seems like some force is being "lost" when I apply it through the block.

If you push the block vertically into the ceiling, then there is no motion. In that case the force does not cause horizontal motion. In your terms, all of the force is "lost".

Force is a vector and can be decomposed into components along any set of directions. A force imparted at an angle is equivalent to two separate horizontal and vertical forces of a certain magnitude, which are determined by the angle.

You should have studied force decomposition in your course somewhere.