# Homework Help: Force and Tension - Calculating Force on another object

1. Oct 11, 2012

### AstoAerial

1. The problem statement, all variables and given/known data
From the text: A farm tractor tows a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer? (Ignore friction)

2. Relevant equations
I'm not sure which are relevant, but here are the ones I have:
ƩF=ma
ƩFy=mg+FN=may
x=.5axt2

3. The attempt at a solution

First, I drew the system; one block pulling another with a string between them, up an incline. I calculated the force of gravity (g) on the trailer: mg=3700*9.81 = 36,297N(newtons) .

I created tilted coordinates and calculated the y force and x force on the trailer Fy1 = 18cos36,297 = 8.172N
Fx1 = 18sin36,297 = -16.038N

I figured that a (acceleration) for the ƩFnet must be 0 since the problem says "steady speed". Constant velocity = zero acceleration, right? But this means that net force on the trailer is also zero. I don't understand this.

I tried first, F=ma, using 3700, but with a=0, F=0. That doesn't work. Then I tried to break down the components and find each force. I only have the components of the trailer, not the tractor, and I feel like I should have some information about the tractor in order to determine force on the trailer, but I'm not sure where to get this information.

That's where I'm stuck. I'm not sure what I'm missing. Any help would be greatly appreciated, since I've been working on this all week and haven't come up with an answer (it is now due tomorrow).

Thank you!

Last edited: Oct 11, 2012
2. Oct 11, 2012

### howie8594

When you do math problems, it's always good to not just get your answer and use it, but rather look at that answer and make sure it makes sense. Ask yourself, does it really make sense that the cosine of 18 times 36,297 is 8.172? It shouldn't make sense because it's way off. It's actually 34, 520. Same with the force parallel to the slope. Look at this again.

You are correct. At constant velocity, the net force is 0. I think you may be confusing force with net force. I'm sure you know the definition, but maybe don't really get it in this context. Basically, in this problem, there is a force pulling the trailer down the hill which is a negative quantity. It's gravity. Then there is another force pulling the trailer up the hill that is a positive quantity. That would be the tractor. So the downward force is negative and the upward force is positive. You see just because it's constant speed doesn't mean there aren't any forces acting on it. It just means the positive force and the negative force add up to zero, and that's what the zero net force is referring to. If you get two forces that cancel out, the result is 0 net force which means the object is either at rest or moving at constant velocity.

First remember that since a = 0, ma = 0. That means the forces equal each other. Secondly, the only force you need is the force acting parallel to the slope due to gravity. You don't need any more components than that.

What makes you think you need more information about the tractor? If similar problem was written saying, "how much force is required out of a person to pull a 50kg crate up a 15 degree slope," would you need to know any information about the person? No. You have enough information here. Imagine a person pulling the load up instead of a tractor.

3. Oct 11, 2012

### AstoAerial

(I'm not sure how to quote things but...)

You said: When you do math problems, it's always good to not just get your answer and use it, but rather look at that answer and make sure it makes sense. Ask yourself, does it really make sense that the cosine of 18 times 36,297 is 8.172? It shouldn't make sense because it's way off. It's actually 34, 520. Same with the force parallel to the slope. Look at this again.

I'm very confused by this, because I'm not sure how you got 34,520. The equation should be 18*cos(36,297). I've tried it several different ways, but I still get 8.171828995... etc. I rounded this to 8.172.

Then, That would be the tractor. So the downward force is negative and the upward force is positive.

This means that the x-component of gravity on the trailer is the opposite of the force the tractor exerts on the trailer, right? Would this then come to F=16.038? or is there another piece I am missing? (The answer in the back of the book says it should be 11kN - 11,000 N. I have no idea how they got this, but I want to understand).

Perhaps T (tension) of the rope? So, F = 0 = -16.038 + T, but then this would mean that the tension is 16.038, which is also wrong. There's definitely a simple concept here that I'm missing, but I can't find it in the book or my notes.

4. Oct 11, 2012

### AstoAerial

(deleted)

Last edited: Oct 11, 2012
5. Oct 11, 2012

### howie8594

You're doing the math wrong. What you're doing is taking the cosine of 36,297 and multiplying it by 18. The correct equation is 36,297(cos 18). That will take the cosine of 18 and multiply it by 36,297.

Yes, the x-component, or the component of gravity acting parallel to the slope, is the opposite of the force of the tractor.

6. Oct 11, 2012

### AstoAerial

OMG how could I not have seen that. Thank you so much. I think I need more coffee!

7. Oct 11, 2012

### howie8594

Haha don't worry about it. It happens.