# Homework Help: Kinematics and Work

1. Dec 22, 2009

### jl1642

I've already made my own attempt at this problem, which was marked incorrect. I'm having trouble determining the correct approach.

1. The problem statement, all variables and given/known data
G: m=50 kg, W= 2.2E3 J, (delta)d=60 m, speed is constant, coefficient kinetic friction = 0.26
R: Force applied, and angle

2. Relevant equations
These are the equations I have:
W = F(delta)d cos(theta)

3. The attempt at a solution

The velocity is constant, so net force must be 0. Therefore, I thought that applied force must be equal to force of friction, or (mu)Fn. This doesn't account for the angle, and is incorrect.

Can anyone offer help? I get the feeling I'm only missing some basic equation. Thanks!

2. Dec 22, 2009

### jegues

Can you give us the whole question you were given?

You're just spitting variables at us it isn't really clear what you're looking for, what you've solved for and what you're having trouble with.

3. Dec 22, 2009

### jl1642

Sure;

Parent is pulling wagon. The combined mass of the wagon and children is 50 kilograms. The adult does 2.2 x 10^3 Joules of work pulling the wagon 60 meters at a constant speed. The coefficient of friction between the two is 0.26.

I've been working with : W = F*d*cos(Theta)

If the speed is constant, then the net force is 0.
Fnet = FA + Ff
FA = - Ff
FA = $$\mu$$ m * g

This approach ignores the angle, and I'm not sure how to proceed.

4. Dec 22, 2009

### jegues

You still haven't told me what you're looking to solve, do you want to find the resultant force?

5. Dec 22, 2009

### jl1642

I want to solve for the force applied by the parent.

6. Dec 22, 2009

### jegues

Okay, for simplicity lets assume he is pulling the wagon to the right.

So W = (Fnet)(d)cos(theta)

We know W, we know d and what do you think the angle will be?

He's pulling it along a flat surface to the right? What angle do we assign to a flat surface?

7. Dec 22, 2009

### jl1642

The following question asks the angle that the wagon is pulled, so the angle is not 0.

I know there are horizontal and vertical components of the applied force, that is what I am trying to find out.

8. Dec 22, 2009

### PhanthomJay

The Normal force is not mg. The vertical component of the applied force must be considered when calculating the Normal force.

9. Dec 22, 2009

### jl1642

I thought that the normal force is exactly equal to the force of gravity, but in the opposite direction. I see what you mean though.

If the force of friction is equal to the x component of FA then F cosθ = (mu) Fn

But how do I get Fn then?

10. Dec 22, 2009

### jl1642

Oh! Do I subtract the y component of applied force from the normal force? Is that what is equal to Fg?

11. Dec 22, 2009

### yamugushi

If you're just solving for the force then it's:
W (joules) = Force * distance, the rest of the info is extraneous.
2.2e3 = F * 60
F=2.2e3/60

12. Dec 22, 2009

### jl1642

No, I know it is more complex than that. Thanks for the help though.

13. Dec 22, 2009

### jl1642

Work only applies for the force applied in the direction of displacement, which is horizontal. It can't help me with the force applied vertically.

14. Dec 22, 2009

### cepheid

Staff Emeritus
You're on the right track, you're just messing up the wording. You subtract the y component of applied force from the weight to get the normal force. The normal force is the net vertical force on the wagon. In this case, more than one thing contributes to it, not only its weight. The upward component of the parent's applied force also matters so that:

normal force = weight - (upward applied force)

The fact that the person is pulling upwards on the wagon means that it is not pressing down with its full weight and hence the ground does not have to support it as much (normal force is reduced).

15. Dec 22, 2009

### jl1642

Thanks!!! I didn't know that normal force was the net upward force.

So mg - FAy = FN

16. Dec 22, 2009

### cepheid

Staff Emeritus
HOWEVER, the problem statement (which you have really failed to impart very clearly) does not mention anything about an angle. For all we know, the parent could be pulling perfectly horizontally, in which case the problem is simpler.

17. Dec 22, 2009

### cepheid

Staff Emeritus
Well yeah, "normal" means "perpendicular to", and the normal force is a contact force that acts perpendicular to any surface an object is in contact with. In the case of a wagon sitting on the ground, the normal force is the force with which the ground pushes upward on the wagon. By Newton's third law, this is equal in magnitude (and opposite in direction) to the force with which the wagon pushes down on the ground. So obviously that's not always going to be the wagon's weight if other vertical forces are acting as well.

18. Dec 22, 2009

### jl1642

As I said, the next question asks what angle the force is applied at, so I know it isn't horizontal. The answer I gave has already been marked, and the correct value for the angle of the force applied was 84.0 degrees.

19. Dec 22, 2009

### jl1642

It wasn't obvious in my course material, but thanks for the help.

20. Dec 22, 2009

### cepheid

Staff Emeritus
Fair enough. You should have two equations (sum of forces = 0 in x and y directions) and two unknowns (the angle theta, and the magnitude of the applied force F). If the number of equations equals the number of unknowns, then there is a unique solution.

Genuinely glad I could help clarify the concept if it wasn't clear before.

21. Dec 22, 2009

### jl1642

Okay, so if the net horizontal and vertical force is 0, this is what I have so far:
$$\Sigma$$Fx = 0 = $$\mu$$ (mg - FAy) + FAx

and FAx = W / d

I think I have it now. Thanks!

22. Dec 22, 2009

### cepheid

Staff Emeritus
Exactly. And you know that for the applied force, the components are related by:

F = Fx / cos(theta)

Fy = Fsin(theta) = Fx [sin(theta)/cos(theta)] = Fx tan(theta)

23. Dec 22, 2009

### jl1642

Thanks so much cepheid, I've been working on this for a while and couldn't get any headway. :-)

24. Dec 22, 2009

### jegues

Sorry if I added confusion but how the question was worded it was safe to assume we were working on a flat surface...