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

[Warmacblu]

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

W_g = mgd

w = Fdcos(theta)

The Attempt at a Solution

I actually figured out part 2 by using W_g = 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.

 

[kuruman]

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.

 

[Warmacblu]

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)?

 

[RoyalCat]

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.

 

[Warmacblu]

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?

 

[RoyalCat]

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.

 

[Warmacblu]

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).

 

[Doc Al]

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?

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.)

 

[RoyalCat]

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 $$\mu N$$
Second, your suggestions for $$N$$ are dimensionally inconsistent (Forces+pure numbers don't add up)

Please, draw out the free body diagram.
Find the net force in the $$x,y$$ 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 $$N$$ and the net force parallel to the wall.

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

 

[Warmacblu]

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

Please, draw out the free body diagram.
Find the net force in the $$x,y$$ 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 $$N$$ and the net force parallel to the wall.

I'll give you a hint about the normal force, it doesn't take $$mg$$ 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.

 

[Doc Al]

Looks good except for mixing up sine and cosine.

 

[Warmacblu]

Okay, I will try solving for F and post the equation.

 

[Warmacblu]

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.

 

[Doc Al]

Looks good except for mixing up sine and cosine.

I take it back. What's with the d's?

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.)

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.

 

[Warmacblu]

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?

 

[Doc Al]

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).

 

[Warmacblu]

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.

 

[Doc Al]

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.

 

[Warmacblu]

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

 

[Doc Al]

How does this look?

Fsin27 - mg - uFcos27 = 0
-uFcos27 = mg - Fsin27

OK.

F = tan27/u + mg

:yuck:

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.

 

[Warmacblu]

OK.

:yuck:

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?

 

[Doc Al]

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.

 

[Warmacblu]

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.

 

[Doc Al]

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?

 

[Warmacblu]

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.

 

[Doc Al]

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.

 

[Warmacblu]

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.

 

[Doc Al]

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.

 

[Warmacblu]

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

 

[Doc Al]

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"

 

[Warmacblu]

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"

I love HyperPhysics, I have been referencing that site since day one.

I got the correct answer for the first part ... 142.99. Now onto the third part.

I am asked to find the magnitude of the normal force between the block and the wall. The normal force is being exerted by the wall to counter the force I put on the block.

The force I put on the block is what I just solved ... Fdcos(theta) and the work done by the normal force is ...

I can't seem to come up with a definite answer but I can take a stab at it.

I believe the normal force would be the opposite of the force put on the block, so -Fdsin(theta)?

I am having a hard time understand the normal force for this problem.

Any help is appreciated.

I just thought about this. The normal force is the same force exerted by the first force from part 1 so I have to convert that number into N to get my answer. But I don't know how that would work if the force is at an angle.

 

[Doc Al]

Part C just asks for the magnitude of the normal force. Go back to your original equation for the horizontal forces, which will tell you N in terms of F (which you already know now). (Make sure you use the proper angle and trig function.) This part has nothing to do with work.

 

[Warmacblu]

Part C just asks for the magnitude of the normal force. Go back to your original equation for the horizontal forces, which will tell you N in terms of F (which you already know now). (Make sure you use the proper angle and trig function.) This part has nothing to do with work.

Alright, my original equation is:

N - Fdsin27

So N = Fdsin27

I know F = 142.99
I know d = 3m

Now about the angle and trig function ... I believe the trig function will be cos because the angle is 27 degrees above the horizontal and the angle will still be 27.

 

[Doc Al]

Alright, my original equation is:

N - Fdsin27

So N = Fdsin27

This is incorrect:
(1) You have the wrong trig function
(2) You are mixing work with force (work is irrelevant here)

Now about the angle and trig function ... I believe the trig function will be cos because the angle is 27 degrees above the horizontal and the angle will still be 27.

This is true, so correct your equation.

 

[Warmacblu]

This is incorrect:
(1) You have the wrong trig function
(2) You are mixing work with force (work is irrelevant here)


This is true, so correct your equation.

Since cos and 27 degrees is correct, I believe N = Fcos27. However the N force is in the negative direction. I did try N = Fcos27 in the positive direction and that answer was incorrect. I just want to make sure that the answer has to be negative before I use any more tries.

Thanks.