# Tractor up a slope

[SOLVED] Tractor up a slope

The question is as follow...
A farm tractor tows a 4300kgs trailer up an incline 26 degrees aboe the horizontal at a steady speed of 3.0m/s. What force does the the tractor exert on the trailer?

I've read the section over in which this problem is included in...
the section is over find the accerlation and force of in 2 dimensions.

well if i'm not mistaken...the accerlation of this problem is 0. F=ma
F=4300(0)=0? Doesn't make sense

OR

if the tractor is pulling this tailer...the tractor is obviously exerting the force of the trailer and then some if it's moving it by 3m/s

Am I going in the rigth direction?

## Answers and Replies

Mentor
Is the acceleration (of the trailer) = 0? YES!

SO... does F = MA = 0? YES!

But you need to understand what F=MA means. It really should be written as:

Fnet = MA

The net force is zero. But there are several forces acting on the trailer. What are they?

Fy and Fx

Mentor
Originally posted by izmeh
Fy and Fx
Good one! Now tell me the real forces. What's pushing or pulling on the trailer? Then we can break them into x and y components.

PrudensOptimus
In the Y direction, the total force is 0.

But in the X Direction, the Force is 18472.960125691722371458233543339 N.

Last edited:
Originally posted by Doc Al
Good one! Now tell me the real forces. What's pushing or pulling on the trailer? Then we can break them into x and y components.

the tractor is pulling on it at the angle...
gravity is pulling it down

Mentor
Originally posted by izmeh
the tractor is pulling on it at the angle...
gravity is pulling it down
Correct. (The incline is also pushing the trailer.)

So... what is the component of the weight along the incline?

-FySin(26) + FxCos(26)?

I don't know

Mentor
The weight of the trailer is (mass)x (g)= mg.

The component of the weight along the incline is mg sin(26) pointing down. (Draw a picture!) That must be balanced by the force of the tractor pulling the trailer up the incline. So...

Forcetractor pulling trailer = mg sin(26)

18472.96n

PrudensOptimus
Originally posted by izmeh
18472.96n

... should've read my response.

Mentor
Originally posted by PrudensOptimus
... should've read my response.
Actually, he should have ignored your response. Just tossing out an "answer" without any explanation is not helpful.

Also, your answer was confusing: "In the Y direction, the total force is 0." Now, what did you mean by that?

Mentor
Originally posted by izmeh
18472.96n
Yep!