# Homework Help: Applied Force

1. Jan 20, 2007

### ND3G

In many neighbourhoods, you might see parents pulling youngsters in a four-wheeled wagon. The child and the wagon have a combined mass of 50kg and the adult does 2.2 x10^3 J of work pulling the two 60m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26

Determine the magnitude of the force applied by the parent

Given:
m = 50kg
W = 2200 J
d = 60m
u = 0.26

Required: Fapp

Analysis: Fapp = Fnet - Ff; Fnet = W/d; Ff = uFn; Fn = Fg; Fg = mg

Solution: Ff = (0.26)(50kg)(-9.8m/s^2) = -127.4 N

Fnet = 2200J / 60m = 36.7N

Fapp = 36.7N - (-127.4N) = 163.4N

Paraphrase: The Magnitude of the force applied by the parent is 163.4N.

Can someone kindly check this over for me and let me know if I have solved the question correctly? I haven't applied (cos) to the formula
W = F*cos*d but I wasn't sure how to solve for that part...

2. Jan 20, 2007

### Decan

Seems right to me. Since the problem does not give you an angle, you can only assume that the applied force is parallel to the horizontal...

3. Jan 20, 2007

### ND3G

Well actually, the next question asks me to determine the angle at which the parent is applying this force. I could state that it is 0 since that is what I used to solve the first part, but that is kind of a self-fulfilling answer.

I feel like I am still missing something.

Last edited: Jan 20, 2007
4. Jan 20, 2007

### Decan

Based on the problem, that would be my best guess...maybe i'm wrong?

5. Jan 20, 2007

### PhanthomJay

If you take your result for the
force applied by the parent, and multiply it by the distance moved, you don't get 2200J, so something is wrong. If the parent does 2200J of work, and the cart moves at constant speed, then how much work is done by friction?
Use that value and go from there.

6. Jan 21, 2007

### ND3G

OK, so here we go farther down the rabbit hole

Required:
Fapp

Analysis:
Since the speed is constant Fnet = 0
Therefore, F(a) - F(f) = 0, F(a) = F(f)
F(f) = uFn, Fn = Fg, Fg = mg

Solution:
Ff = (0.26)(50kg)(-9.8m/s^2) = -127.4 N

Analysis:
W = F(cos(theta))(d), F(cos(theta)) = W/d

Solution:
F(cos(theta)) = 2200J / 60m = 36.7 N

Analysis:
Since the parent is pulling both forward and upwards:
Fn = -[F(sin(theta)) + mg]
Pulling Force + Frictional Force = 0
Therefore, F(cos(theta)) - (0.26)[F(sin(theta)) + mg] = 0

Solution:
36.7 N - (0.26)[F(sin(theta)) - 127.4] = 0
141.2 N - [F(sin(theta)) - 127.4 N] = 0
F(sin(thetha)) = 268.6 N

Analysis:
sin(theta) / cos(theta) = tan(theta)
Cancel out F:

Solution:
268.6 N / 36.7 N = 7.31880108992

Theta = tan^-1 (7.31880108992) = 82.2 degrees

Analysis:
F = W / ((cos(theta))(d))

Solution:
F = 2200 J / ((cos(82.2)(60m)) = 270.2 N

Paraphrase:
The magnitude of force applied by the parent is 270.2 N

I'm almost positive that I still don't have it right but I think I am getting closer

Last edited: Jan 21, 2007
7. May 3, 2007

### nida

That's not the way you do it. U're still incorrect. The Fn is not = Fg. if you still need help with the question let me know.