What is the Work and Normal Force in a Block Pushed Up a Wall?

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Homework Help Overview

The problem involves a 2.0 kg block being pushed up a vertical wall at a constant velocity by a force applied at an angle of 27 degrees. The discussion centers around calculating the work done by this force, the work done by gravity, and the normal force acting on the block, with a coefficient of kinetic friction of 0.30.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the necessity of drawing a free body diagram to identify all forces acting on the block. There are attempts to relate the work done to the force and distance, with some questioning the assumptions about the force being zero. Others express confusion about the direction of the normal force and its relationship to the weight of the block.

Discussion Status

The discussion is ongoing, with participants sharing insights about the relationships between forces and the need for accurate equations. Some have drawn free body diagrams and are attempting to clarify their understanding of the forces involved, while others are correcting misconceptions about the normal force and its calculation.

Contextual Notes

Participants are navigating the complexities of the problem, including the definitions of work and the roles of different forces. There is an emphasis on ensuring that all terms in equations are dimensionally consistent, and some participants are grappling with algebraic manipulations related to the forces involved.

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Homework Statement



A 2.0 kg block is pushed 3.0 m at a constant velocity up a vertical wall by a constant force applied at an angle of 27 degrees with the horizontal.

The acceleration of gravity is 9.81 m / s2.

If the coefficient of kinetic friction between the block and the wall is .30 ...

A) Find the work done by the force on the block. Answer in units of J.

B) Find the work done by gravity on the block. Answer in units of J.

C) Find the magnitude of the normal force between the block and the wall. Answer in units of N.

Homework Equations



Wg = mgd

w = Fdcos(theta)

The Attempt at a Solution



I actually figured out part 2 by using Wg = mgd

I just can't seem to figure out how to set up parts one and three. I think I have to use w = Fdcos(theta) and set that equal to the force of gravity for part one. And for part three I have to use uN which will equal umg.
 
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You cannot do this problem correctly unless you draw a free body diagram and put in all the forces. When you do that, you will see that the normal force is not mg. "Normal" means "perpendicular to the surface". Here the weight mg is parallel to the surface.
 
Okay, I drew a free body diagram and I see what you mean which makes this question seem even more confusing. I believe I am trying to solve for the force between the block and the wall but the normal force is not even in that direction.

Also, for part 1:

I believe I need to use w = Fdcos(theta) but since I am not given a force and it is constant, can I assume that it is just zero? So I would then have w = dcos(theta)?
 
Warmacblu said:
Okay, I drew a free body diagram and I see what you mean which makes this question seem even more confusing. I believe I am trying to solve for the force between the block and the wall but the normal force is not even in that direction.

Also, for part 1:

I believe I need to use w = Fdcos(theta) but since I am not given a force and it is constant, can I assume that it is just zero? So I would then have w = dcos(theta)?

The normal force is the force between two objects in contact, perpendicular to the surface of their contact. It is NOT the force that opposes gravity.

Draw the free body diagram, and remember that the normal is between the mass and the wall.
 
That is what I first thought. But before I answer the third part I would really like to figure out the first part. Is my assumption correct and would w = dcos(theta) actually work?
 
Warmacblu said:
That is what I first thought. But before I answer the third part I would really like to figure out the first part. Is my assumption correct and would w = dcos(theta) actually work?

Yes, that would be perfectly true. That's just the definition of work.
You have to pay close attention to what the net force parallel to the wall is, though, or you'll end up making the same mistakes.
 
So the net force parallel to the wall for the first problem would be uN. I am not sure whether N is mg - sin(theta) or mg - cos(theta).
 
Warmacblu said:
Also, for part 1:

I believe I need to use w = Fdcos(theta) but since I am not given a force and it is constant, can I assume that it is just zero?
:confused: Why do you think you can just assume that the force is zero? Draw a free body diagram and figure out what the force is.
So I would then have w = dcos(theta)?
No. If you assume that F = 0, then w = 0. (w = dcos(theta) assumes F = 1, not F = 0.)
 
Warmacblu said:
So the net force parallel to the wall for the first problem would be uN. I am not sure whether N is mg - sin(theta) or mg - cos(theta).

This is also false.
First, the force parallel to the wall is not just [tex]\mu N[/tex]
Second, your suggestions for [tex]N[/tex] are dimensionally inconsistent (Forces+pure numbers don't add up)

Please, draw out the free body diagram.
Find the net force in the [tex]x,y[/tex] axes. What you are allowed to assume is that the block doesn't move in the x-axis (That wouldn't make much sense either)
Use NSL to find the value of [tex]N[/tex] and the net force parallel to the wall.

I'll give you a hint about the normal force, it doesn't take [tex]mg[/tex] in as a factor.
 
  • #10
RoyalCat said:
This is also false.
First, the force parallel to the wall is not just [tex]\mu N[/tex]
Second, your suggestions for [tex]N[/tex] are dimensionally inconsistent (Forces+pure numbers don't add up)

Please, draw out the free body diagram.
Find the net force in the [tex]x,y[/tex] axes. What you are allowed to assume is that the block doesn't move in the x-axis (That wouldn't make much sense either)
Use NSL to find the value of [tex]N[/tex] and the net force parallel to the wall.

I'll give you a hint about the normal force, it doesn't take [tex]mg[/tex] in as a factor.

Okay, I drew out my free body diagram again and came up with this:

Forces in the horizontal:

N - Fdsin27

Forces in the vertical:

Fdcos27 - mg - uN

So N = Fsin27 and the forces in the vertical will equal:

Fdcos27 - mg - uFdsin27

I guess the next step would be to solve for F, but before I go on I want to make sure I am correct in my force equations.

Thanks for the help.
 
Last edited:
  • #11
Looks good except for mixing up sine and cosine.
 
  • #12
Okay, I will try solving for F and post the equation.
 
  • #13
Fdsin27 - mg - uFdcos27 = 0
-uFdcos27 = -Fdsin27 + mg
F = utan27 + mg

I am not sure if I can divide out the d or if the F's would end up on the left side. My algebra is a bit weak.
 
  • #14
Doc Al said:
Looks good except for mixing up sine and cosine.
I take it back. What's with the d's?
Warmacblu said:
Okay, I drew out my free body diagram again and came up with this:

Forces in the horizontal:

N - Fdsin27

Forces in the vertical:

Fdcos27 - mg - uN
In addition to the sine/cosine mixup, you have extraneous d's stuck in there. Get rid of them. (Sorry for not spotting that earlier.)

Warmacblu said:
Fdsin27 - mg - uFdcos27 = 0
-uFdcos27 = -Fdsin27 + mg
F = utan27 + mg

I am not sure if I can divide out the d or if the F's would end up on the left side. My algebra is a bit weak.
The d should not be there. For example, the vertical component of F is Fsin27, not Fdsin27. Don't mix up force and work. Here you're just dealing with forces.
 
  • #15
Doc Al said:
I take it back. What's with the d's?

In addition to the sine/cosine mixup, you have extraneous d's stuck in there. Get rid of them. (Sorry for not spotting that earlier.)


The d should not be there. For example, the vertical component of F is Fsin27, not Fdsin27. Don't mix up force and work. Here you're just dealing with forces.

The first part of the problem asks for work done by the force. I thought that in order to solve for work, a distance has to be taken into account?
 
  • #16
Warmacblu said:
The first part of the problem asks for work done by the force. I thought that in order to solve for work, a distance has to be taken into account?
That's true, but first you need to solve for the force. Then you can worry about the work done.

Also: in any physically meaningful equation, every term must have the same units. You can't have a work term (Fd) added to a force term (mg).
 
  • #17
Doc Al said:
That's true, but first you need to solve for the force. Then you can worry about the work done.

Also: in any physically meaningful equation, every term must have the same units. You can't have a work term (Fd) added to a force term (mg).

Okay, so I plug my numbers into F = utan27 + mg and then use that F with W = Fd? Where d is equal to 3 m.
 
  • #18
Warmacblu said:
Okay, so I plug my numbers into F = utan27 + mg
That equation isn't correct; redo your algebra in solving for F. (Note that the "utan27" term has no units, thus cannot be added to a force term.)
and then use that F with W = Fd? Where d is equal to 3 m.
Don't forget to take the angle into account. W = Fd only if F and d are in the same direction.
 
  • #19
Doc Al said:
That equation isn't correct; redo your algebra in solving for F. (Note that the "utan27" term has no units, thus cannot be added to a force term.)

Don't forget to take the angle into account. W = Fd only if F and d are in the same direction.

How does this look?

Fsin27 - mg - uFcos27 = 0
-uFcos27 = mg - Fsin27
F = tan27/u + mg
 
  • #20
Warmacblu said:
How does this look?

Fsin27 - mg - uFcos27 = 0
-uFcos27 = mg - Fsin27
OK.
F = tan27/u + mg


Just put all terms containing F on one side, then isolate F.

Can you solve this equation?:
ax -c -bx = 0
If so, it's the same thing here.
 
  • #21
Doc Al said:
OK.



Just put all terms containing F on one side, then isolate F.

Can you solve this equation?:
ax -c -bx = 0
If so, it's the same thing here.

Putting it that way makes it easier for me.

ax - c - bx = 0
ax - bx = c
x(a-b) = c
x = c / (a-b)

F = mg / (ucos27 - sin27)

How does that look?
 
  • #22
Warmacblu said:
F = mg / (ucos27 - sin27)

How does that look?
Much better. But you flipped the order of the terms in the denominator, thus your answer will have the wrong sign. Fix that.
 
  • #23
Doc Al said:
Much better. But you flipped the order of the terms in the denominator, thus your answer will have the wrong sign. Fix that.

Okay, I swapped the bottom terms and got 105.095. From there I am going to solve for W. If W = Fd, when F is 105.095, d is 3cos27.
 
  • #24
Warmacblu said:
Okay, I swapped the bottom terms and got 105.095.
Check your arithmetic.
If W = Fd, when F is 105.095, d is 3cos27.
No. What's the angle between the force and the displacement?
 
  • #25
Doc Al said:
Check your arithmetic.

No. What's the angle between the force and the displacement?

I don't know how I came up with 105.095 but I am now coming up with 104.988. The angle between the force and the displacement is sin27 but I thought since the distance was moving in the y direction, it has to be in terms of cos.
 
  • #26
Warmacblu said:
I don't know how I came up with 105.095 but I am now coming up with 104.988.
Still not right. Show the equation you're using and the numbers you are plugging in.
The angle between the force and the displacement is sin27 but I thought since the distance was moving in the y direction, it has to be in terms of cos.
The angle that the force makes with the horizontal is 27 degrees. But the displacement is vertical.
 
  • #27
Doc Al said:
Still not right. Show the equation you're using and the numbers you are plugging in.

The angle that the force makes with the horizontal is 27 degrees. But the displacement is vertical.

I used F = (mg) / (sin27 - ucos27)

F = (2 * 9.8) / (sin27 - .30cos27)

F = 19.6 / .187

F = 104.99

The angle the force makes with the displacement vector would be 27 + 90 = 117.
 
  • #28
Warmacblu said:
I used F = (mg) / (sin27 - ucos27)

F = (2 * 9.8) / (sin27 - .30cos27)

F = 19.6 / .187

F = 104.99
My bad--you were correct.

The angle the force makes with the displacement vector would be 27 + 90 = 117.
No. The force is 27 degrees above the horizontal and the displacement vector (which is vertical) is 90 degrees above the horizontal. So the angle between them is 90-27.
 
  • #29
Doc Al said:
My bad--you were correct.


No. The force is 27 degrees above the horizontal and the displacement vector (which is vertical) is 90 degrees above the horizontal. So the angle between them is 90-27.

Okay, I understand that. But I don't know for sure if it's sin or cos. Could you explain that a little further?

W = 104.99 * 3sin63 = 280.64

W = 104.99 * 3cos63 = 142.99
 
  • #30
Work = F*d*cosθ, where θ is the angle between the force and the displacement. Note that F*cosθ can be viewed as the component of force in the direction of the displacement. If there's no component of force in the direction of the displacement, then that force does no work.

This might help a bit: http://hyperphysics.phy-astr.gsu.edu/hbase/work2.html#wdef"
 
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