Forces on a crate with friction: find the normal force

In summary: And yes I understand the purpose of the problem. In summary, the conversation is about solving a physics problem involving a normal force and a vertical net force of zero. The conversation includes a discussion about the purpose of the problem and a clarification on plugging in the correct values for force P and kinetic frictional force f. The final answer is 104.79 N.
  • #1
Jujubee37
22
4
Homework Statement
Force P (78.28 N) pulls on a crate of weight W (164 N) on a rough surface at a constant speed. The figure shows the directions of the forces that act on the crate. FN represents the normal force. The kinetic frictional force is f (51.25 N). The angle, a, is 49.1 degrees. Calculate the value of the normal force.
Relevant Equations
mg-Fsin(x)
1613252983020.png
I have attempted this problem by solving for the normal force. (16.73)(9.8)-51.25(sin)(49.1). I tried to work that out but it was incorrect because apparently the vertical force is zero. Could I get an explanation on what that means and where I should start?
 
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  • #3
I did. (16.73)(9.8)-51.25(sin)(49.1)= 125.21 N
 
  • #4
Are these even in the same direction ?
 
  • #5
BvU said:
Are these even in the same direction ?
I wouldn't know. All I know is that the vertical net force is zero and the normal force is one of the vertical forces. I worked this out two ways. one with sin and the other with cos because those are the only equations that I have involving an angle.
 
  • #6
Jujubee37 said:
I wouldn't know.
Huh ? Isn't mg vertical and 51.25 N horizontal ?

Jujubee37 said:
apparently the vertical force is zero
Says who ?
 
  • #7
BvU said:
Huh ? Isn't mg vertical and 51.25 N horizontal ?

Says who ?
Lon capa the website. I don't understand this so I can't answer your questions.
 
  • #8
I can't get in there, so you have to play go-between.

Is it clar to you that ##\ mg\ ##, ##\ \ P\sin a\ \ ## and ## \ N \ ## are the only vertical forces in the story ?
And if ##N## would be zero, then ##\ mg = P\sin a\ \ ## would follow ?

Just to humor me :wink: , calculate the net horizontal force ...

##\ ##
 
  • #9
BvU said:
I can't get in there, so you have to play go-between.

Is it clar to you that ##\ mg\ ##, ##\ \ P\sin a\ \ ## and ## \ N \ ## are the only vertical forces in the story ?
And if ##N## would be zero, then ##\ mg = P\sin a\ \ ## would follow ?

Just to humor me :wink: , calculate the net horizontal force ...

##\ ##
I still do not understand what I have to do. Solve for the horizontal? or what?
 
  • #10
Why not just do it ?

##N## can't be zero: ##mg## is much bigger than ##P\sin a##.
 
  • #11
BvU said:
Why not just do it ?

##N## can't be zero: ##mg## is much bigger than ##P\sin a##.
Do what? I really do not understand where I am supposed to start. I have no understanding of this problem. This is what the hint was for this problem.

The net vertical force here is zero. The normal force is one of the vertical forces. Others are weight and part of the pulling force.
 
  • #12
Aha ! It says the NET vertical force is zero. There is no acceleration in the vertical direction. So you have ##mg## down, ##P\sin a## and ##N## both up. And the sum is zero.

The only difference with your working is that you used 51.25 instead of 78.28 N for ##P##.
 
  • #13
Jujubee37 said:
I did. (16.73)(9.8)-51.25(sin)(49.1)= 125.21 N
Where that 16.73 comes from?
 
  • #14
I guess it's a calculated ##m##
 
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  • #15
Thank you, BvU; exactly.
Perhaps not being at scale, the posted diagram is confusing the OP.
 
  • #16
However, I suspect @Jujubee37 is still confused ?
 
  • #17
Jujubee37 said:
... I really do not understand where I am supposed to start. I have no understanding of this problem.

Not being pulled in any direction, the block pushes the ground down with force W of magnitude 164 N and there is no kinetic frictional force f to talk about.
Do you understand that?
 
  • #18
i got the answer already. 16.73 is the mass. I had to solve for it
 
  • #19
Jujubee37 said:
i got the answer already. 16.73 is the mass. I had to solve for it
May I ask what answer you got?
Do you now understand the purpose of this problem?
 
  • #20
@Jujubee37: You do not need to find the mass. Simply replace 'mg' by '164' (Newtons).

Have you solved the problem now, or do you still want help?
 
  • #21
Jujubee37 said:
Homework Statement:: Force P (78.28 N) ... The kinetic frictional force is f (51.25 N).

View attachment 277913
. (16.73)(9.8)-51.25(sin)(49.1).
Looks to me that you got confused between P and f when plugging numbers into your equation.
 
  • #22
Lnewqban said:
May I ask what answer you got?
Do you now understand the purpose of this problem?
The answer I got was 104.79 N. I have practiced more problems similar to this one so I understand it now.
 
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  • #23
Steve4Physics said:
@Jujubee37: You do not need to find the mass. Simply replace 'mg' by '164' (Newtons).

Have you solved the problem now, or do you still want help?
Yes I am aware now, my way got me the correct answer but yes plugging in 164 would be quicker.
 

1. What is a normal force?

The normal force is the force that a surface exerts on an object that is in contact with it. It is always perpendicular to the surface and is equal in magnitude but opposite in direction to the force exerted by the object on the surface.

2. How is the normal force related to forces on a crate with friction?

The normal force plays a crucial role in determining the amount of friction between the crate and the surface it is on. It is directly proportional to the force of friction, meaning that as the normal force increases, so does the force of friction.

3. Why is friction important when considering forces on a crate?

Friction is important because it is the force that opposes the motion of the crate. Without friction, the crate would slide or move easily, making it difficult to control or transport. Friction helps to keep the crate in place and allows for better control.

4. How does the coefficient of friction affect the normal force?

The coefficient of friction is a measure of the roughness of the surfaces in contact. It affects the normal force by determining the amount of friction that will be present. A higher coefficient of friction means a higher normal force and therefore a greater force of friction.

5. How can the normal force be calculated for a crate with friction?

The normal force can be calculated by using the equation: N = mg + Ff, where N is the normal force, m is the mass of the crate, g is the acceleration due to gravity, and Ff is the force of friction. The normal force can also be calculated by resolving the forces in the vertical direction and setting them equal to zero.

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