What is the acceleration of the cart? :)

1. Apr 20, 2017

Dusty912

1. The problem statement, all variables and given/known data
A man is pulling himself up the 15 degree incline by the method shown. If the combined mass of the man and the cart is 100kg, determine the acceleration of the cart if the man excerpts 250 N on the rope. There is also a .2 coefficient for the kinetic friction between the wheels and the ground.

2. Relevant equations

F=m*a
Fk=μ*N
3. The attempt at a solution
So I pretty much know how to solve this without friction being involved. Where T is the friction in the rope, m is the mass of the man and cart theta is the angle, and a is acceleration.

F=m*a
4T-m*g*sinθ=m*a

now when you throw friction into the mix should it look like this?
F=m*a
4T-m*g*sinθ-μ*N=m*a
4T-m*g*sinθ-μ*m*g*cosθ=m*a

with the numerical values
4*(250)-100*9.81*sin(15)-.2*100*9.81*cos(15)=100*a
a=5.566m/2^2

Thanks ahead of time, you all rock

2. Apr 20, 2017

TomHart

Is there a figure you can attach?

3. Apr 20, 2017

haruspex

Sack the question setter. Presumably the wheels are rolling, so kinetic friction between them and the ground is irrelevant, and static friction would not cost any energy. It should say "rolling resistance", not friction. (0.2 is rather high.)
You mean tension, I assume.
I infer that the pulley system involves four lengths of rope between the cart and the top of the slope. If so, your method and answer look fine.

4. Apr 20, 2017

Dusty912

5. Apr 20, 2017

Dusty912

yea, tension in the rope not friction

6. Apr 20, 2017

Dusty912

and are you saying the question is worded incorrectly? because that was a word for word question by professor wrote

7. Apr 20, 2017

haruspex

8. Apr 20, 2017

Dusty912

okay well thanks for pointing that out. we have a forum for our class, so I'll let others know.