Solve Incline Question: 100N Sled, Coefficient of Friction 0.2

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Homework Help Overview

The discussion revolves around a physics problem involving a boy dragging a 100N sled up a 20-degree slope at a constant velocity, with a coefficient of friction of 0.2. Participants are exploring the forces acting on the sled and the implications of the pulling force's angle with respect to the slope.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the meaning of "respect to the hill" and the setup of free body diagrams (FBDs). There are attempts to clarify the components of gravitational force and normal force, as well as the direction of the pulling force. Some participants express uncertainty about their FBDs and the relationships between the forces involved.

Discussion Status

The discussion is active, with participants providing guidance on the interpretation of forces and their components. There is a focus on ensuring that all forces are accounted for and correctly represented in the FBDs. Some participants are questioning their understanding and seeking clarification on the relationships between the forces, particularly regarding the normal force and its dependence on the pulling force.

Contextual Notes

Participants are working under the constraints of the problem statement, which specifies that the sled is being dragged at a constant velocity, implying that the net force is zero. There are ongoing discussions about the correct representation of forces and the implications of the angles involved.

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Homework Statement


A boy drags a 100N sled up a 20degree slope at constant velocity. If the coefficient of friction between sled and hill is 0.2, what force must he exert at an angle of 35 degrees with respect to the hill?


Homework Equations


What does 'respect to the hill' mean?


The Attempt at a Solution


Im not sure where to start? I drew a FBD: perpendicular to slope I have mgcostheta going down and normal force going up perpendicular to slope. F is going to the right and fs is going down to the left. Is this correct?
 
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mizzy said:
What does 'respect to the hill' mean?


The Attempt at a Solution


Im not sure where to start? I drew a FBD: perpendicular to slope I have mgcostheta going down and normal force going up perpendicular to slope. F is going to the right and fs is going down to the left. Is this correct?
(a) respect to slope surface, I guess

(b) I'm not sure if I imagine well your image (what is F?). Can you attach it?

regards
 
F is the force exerted. The boy is dragging the sled up the hill.
 
mizzy said:
F is the force exerted. The boy is dragging the sled up the hill.
Normal force is OK.

F is going "up to the right" with 35 degrees angle respect to the sled surface (it does mean 55 deg to the horizon).

ragards
 
Hello Mizzy,

Welcome to Physics Forums!

mizzy said:
Im not sure where to start? I drew a FBD: perpendicular to slope I have mgcostheta going down
You should reevaluate that force. You will have the gravitational force, of course, but this force should point really straight down (i.e. not perpendicular to the slope). What do you think about its magnitude?
and normal force going up perpendicular to slope.
Okay, that's good. :approve: What's its magnitude, by the way?
F is going to the right
Hmmm. I guess it depends on what you mean by "right"

The slope itself is 20o from the horizontal, and the boy is pulling at an additional 35o with respect to that, making the pulling angle 55o with respect to the horizontal. (I think this answers your first question.)
and fs is going down to the left.
The direction of that last one sounds okay to me, I think. Do you mean that Fs is perpendicular to the surface?
 
yes gravitational force is going straight down, but y component is perpendicular to slope.

Normal force is 100N, right?

Yes, Fs is perpendicular to the surface.

I need to find acceleration in this question, right?
 
mizzy said:
yes gravitational force is going straight down, but y component is perpendicular to slope.

Normal force is 100N, right?
Sorry, 100 N is the gravitational force that points straight down (really straight down -- not perpendicular to the slope's surface). You'll have to combine that with a trigonometric function to find the part of the normal force's magnitude associated with gravity.

But be careful! The gravitational force and the angle of the slope are not the only things which affect the normal force's magnitude. There is something else too.
Yes, Fs is perpendicular to the surface.

I need to find acceleration in this question, right?

The problem statement says, "boy drags ... sled ... at constant velocity." So that implies... :wink:
 
ok. I've found acceleration and I got 6.9m/s^2. This is using the angle 55 degrees.

Is that correct?
 
mizzy said:
ok. I've found acceleration and I got 6.9m/s^2. This is using the angle 55 degrees.

Is that correct?

Sorry, no. The problem statement says that the boy and sled are moving at a constant velocity. This directly implies that the acceleration is zero. (If the acceleration was not zero, the velocity would change as a function of time.)

Given that, here is a hint: First find all the forces involved in the problem (three of these individual forces are dependent on the magnitude of the pulling force F). Then perform the vector sum of all the forces and set that equal to zero (don't forget that you're summing vectors and not scalars). Solve for the magnitude of the pulling force F (you already know its direction).
 
  • #10
ok. tell me if i got all the forces right. Normal force, gravitational force, mg cos theta, mg sin theta, force of friction, the x and y components of the pulling force.

is that right?
 
  • #11
mizzy said:
ok. tell me if i got all the forces right. Normal force, gravitational force,
Sounds good so far. :approve:
mg cos theta, mg sin theta,
What forces are these? I don't understand.
force of friction, the x and y components of the pulling force.
Yes, those are fine. But realize that all the forces have both x and y components. It just so happens that the x component of the gravitational force is 0, but the x component still exists (it just happens to be 0). All the other forces have non-zero x and y components.
is that right?

I only see 4 forces. Consider trying it this way (keep in mind that each of these forces has both x and y components):

o Pulling force F. The magnitude of this force is what you are looking for. The direction has already been determined.
o Gravitational force mg. We've established that this force points straight down.
o Normal force Fn. This force points perpendicular to the slope (which is in a partially upward direction). The magnitude of the normal force is influenced by the slope angle, the gravitational force, and the pulling force (the pulling force F is not parallel to the slope's surface, so it has an effect on how hard the sled pushes against the slope -- thus it affects the magnitude of the normal force).
o Frictional force Ff. Points parallel to the slope's surface, which is in a somewhat downward direction. It is a function of the magnitude of the normal force Fn and the friction coefficient. Since the magnitude of Fn is a function of the F, the magnitude of the frictional force is also (indirectly) a function of F.

Do you see how I come up with that?
 
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  • #12
ok, i think I'm lost.

Maybe my FBD is different from yours. I broke the gravitational force into its x and y components.
 
  • #13
mizzy said:
ok, i think I'm lost.

Maybe my FBD is different from yours. I broke the gravitational force into its x and y components.

Theoretically you can do that; i.e. rotate everything 20o, or some other angle, including gravity and everything. You'll get the same answer, but it won't necessarily make make the problem any easier. It might even make things more complicated. But again, you could do that if you want. Doing so doesn't change the magnitudes of the vectors or their relative directions (relative to each other), which is why it is okay to do that.

[Edit: My point is that rotating or not, you still have 4 forces: Pulling, Gravity, Normal, and Friction. Their relative directions are fixed, and essentially given in the problem statement.]

As a rule of thumb, I try to keep gravity pointing in the negative y direction unless there is a compelling reason to do otherwise. But maybe that's just me.

So anyway, (rotating or not) what expression did you get for the magnitude of the normal force (as I mentioned before, this should be a function of F)?

[Edit: is it possible for you to post an image of your FBD?]
 
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  • #14
For normal force, i have N = mg cos theta

i tried attaching the image but it was too big. I'll try again.
 
  • #15
ok here is my diagram.
 

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  • #16
mizzy said:
For normal force, i have N = mg cos theta

There's more to it than that. Remember, there is a component of the pulling force pulling up on the sled, in a direction away from the surface.

Let me describe a different situation (we can go back to the original problem later). For a moment, imagine a block sitting on a bathroom scale (and the scale might be sitting on a sloped surface). The bathroom scale is measuring the normal force. At this point, the scale is registering mgcosθ. But now grab the block and lift up on it. How does that affect the scale's reading? :wink:

[Edit: 'Just looked at your FBD, and it looks fine. :approve:]
 
  • #17
The scale's reading will be lighter.
 
  • #18
mizzy said:
The scale's reading will be lighter.

Yes, :approve:. Now back to the original problem.

The normal force is the sum of the parallel components of all the other forces (and in the opposite direction of this sum).

You got one of them. mgcosθ is the part of the gravitational force that is parallel to the normal force.

The frictional force has no part that is parallel to the normal force (because the normal force and frictional force are perpendicular to each other).

But there is a part of the pulling force F, that is parallel to the normal force that needs to be considered too. This needs to be part of the sum that makes up the normal force.

(By the way, this might be a good time to try out that rotation idea, at least in your mind, for this particular step. Imagine rotating the whole figure such that the normal force points straight up [you can rotate it back later -- this is just to find the normal force]. Now the normal force points straight up, in the positive y direction. It's x component is zero. Find the y components of all the vectors [including the normal force] and set the sum equal to zero.)

You'll find that the normal force has an mgcosθ which you've already found, and also another term which is a function of the pulling force F.
 
  • #19
ok. I solved for F, and I got 75.7N. Is that correct?

I really appreciate your help. Thanks so much.
 
  • #20
mizzy said:
ok. I solved for F, and I got 75.7N. Is that correct?

Sorry, but I double checked and 75.7 N is not the right answer (magnitude of the pulling force). :frown:

If you show your work maybe I can help figure out what went wrong. Start with your expression for the normal force, since that's the first step in solving this problem (remember, you should expect that your expression for the normal force is some function of the pulling force F). :smile:
 
  • #21
collinsmark said:
Sorry, but I double checked and 75.7 N is not the right answer (magnitude of the pulling force). :frown:

If you show your work maybe I can help figure out what went wrong. Start with your expression for the normal force, since that's the first step in solving this problem (remember, you should expect that your expression for the normal force is some function of the pulling force F). :smile:

I went through my calculations again and I got 57N instead of 75.7N.

I summed up all the x and y components:

Sum of Fx = 0 Sum of Fy = 0
Fcos35 - muN - mgsin20 = 0 N - mgcos20 + Fsin35 = 0
therefore, N = mgcos20 - Fsin35

I plugged N into the first equation. Then solved for F.

is that correct?
 
  • #22
mizzy said:
therefore, N = mgcos20 - Fsin35

Yeh! :biggrin: Perfect! :approve:

I went through my calculations again and I got 57N instead of 75.7N.

Great! :approve: You might want to be careful with your precision (significant digits) though. I got something closer to 56.7 N. Still, great job! :smile:
 
  • #23
Thanks so much! :biggrin:
 

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