Solve Kinematics & Work: Find Force & Angle

[jl1642]

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

Homework Statement

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

Homework Equations

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

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!

 

[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.

 

[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.
F_net = F_A + F_f
F_A = - F_f
F_A = \mu m * g

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

 

[jegues]

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

 

[jl1642]

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

 

[jegues]

Okay, for simplicity let's 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?

 

[jl1642]

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

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.

 

[PhanthomJay]

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

 

[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 F_A then F cosθ = (mu) Fn

But how do I get Fn then?

 

[jl1642]

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

 

[yamugushi]

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.
F_net = F_A + F_f
F_A = - F_f
F_A = \mu m * g

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

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

 

[jl1642]

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

 

[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.

 

[cepheid]

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

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).

 

[jl1642]

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

So mg - F_Ay = F_N

 

[cepheid]

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.

 

[cepheid]

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

So mg - F_Ay = F_N

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.

 

[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.

 

[jl1642]

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.

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

 

[cepheid]

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.

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.

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

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

 

[jl1642]

Okay, so if the net horizontal and vertical force is 0, this is what I have so far:
\SigmaF_x = 0 = \mu (mg - F_Ay) + F_Ax

and F_Ax = W / d

I think I have it now. Thanks!

 

[cepheid]

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)

 

[jl1642]

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

 

[jegues]

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