# Acceleration Caused by some Forces on a Cart, Help Please

#### SpeeDFX

This is my first physics class and I'm looking to do more than just pass it, I want to understand everything fully. Anyway, my class just started going over Newtons first, second and third laws, and I'm having trouble with a problem in my book (this problem http://img290.imageshack.us/img290/9241/physhelp1gi.jpg [Broken]. I've scanned both the question, and the solution from the solution manual. When I think about this problem, I must be over simplifying it, because for part (a), I thought the tension would just be the weight of M2 minus the force of m1, but it doesn't seem that way. As for parts (b), (c), and (d), I'm completely lost. This problem is not for credit. I asked my teacher for help, but I still didn't understand him. I'm hoping someone here can explain this to me with words and math rather than just math, because it's easier for me to understand it that way. Thanks.

- Chris

Last edited by a moderator:
Related Introductory Physics Homework News on Phys.org

#### Doc Al

Mentor
SpeeDFX said:
When I think about this problem, I must be over simplifying it, because for part (a), I thought the tension would just be the weight of M2 minus the force of m1, but it doesn't seem that way.
I don't know what you mean by the "force of m1".

In any case, the way to solve these kinds of problems is apply Newton's 2nd law (F = ma) to each object, then solve the resulting set of equations together. (Don't try any shortcuts until you have lots of practice.)

That's what the book's solution is. The three equations given represent Newton's 2nd law applied to the three bodies in this problem: m1, m2, and the cart. Start by identifying all the forces and accelerations involved and see if you can duplicate those three equations on your own.

#### Maxos

Yes, it is a non-banal example of Newton's law working as an algebraic eqation, rather than an DE, as usual.

I understand there is some problem in thinking of the tension of strings, probably it would be better not to draw them entirely, but only spot the ends and thus insert the Force keeping the string of the same length, IE, Tension.

#### SpeeDFX

The process of identifying the forces and accelerations is where I'm having the most trouble. this is where I need help. More specifically, I dont understand how the acceleration of the cart M arises. If it moves to the left, there must be a force going to the left, right? When I look at the pulley, should I treat it as a particle with a horizontal force going to the left of magnitude T, and another force pointing straight down with a magnitude of T also? In the question, what is the significance of the statement "asume that m2 can move only vertically"?
I've been working on this problem for many hours, and something here is just not clicking for me :- \

#### Doc Al

Mentor
SpeeDFX said:
The process of identifying the forces and accelerations is where I'm having the most trouble. this is where I need help. More specifically, I dont understand how the acceleration of the cart M arises. If it moves to the left, there must be a force going to the left, right?
When I look at the pulley, should I treat it as a particle with a horizontal force going to the left of magnitude T, and another force pointing straight down with a magnitude of T also?
Realize that the pulley and cart can be considered a single object. (They are attached.) Since the surfaces are frictionless, the only horizontal force acting on the pulley-cart is the string which pulls to the left with its tension, T.

In the question, what is the significance of the statement "asume that m2 can move only vertically"?
You can assume that the lower portion of the string is vertical. (If m2 accelerates horizontally, the string would have to make an angle with the vertical.)

#### SpeeDFX

Thanks, that helped a lot. I now understand 2 of the 3 equations. I worked out the 2nd one and the 3rd one by myself, but now I'm a bit confused as to how the acceleration "a" of the 2 masses connected by the string is related to the acceleration "A" of the entire cart M.

This is my thought process...
The tension T acts on all 3 objects, m1, m2, and M. If I just look at the the horizontal accelerations, then the tension T causes m1 to accelerate to the right, and also causes M to accelerate to the left. The magnitude of both the force T onto m1 and M is kindof...split.. so that the the magnitude of the force required to move m1 PLUS the magnitude of the force required to move M is T. That makes me think of the following equation...

(m1)(a) + (M)(A) - T = 0, but I think that's wrong. According to the equation (1) in the solution, I'm supposed to relate m1 with A, but I'm not able to grasp this relation.

Sorry for writing so much on a seemingly simple problem. I dont know why this one is so difficult for me.

#### Maxos

Why do you think the force moving M (M*A) applies to m1, too? There is no friction.

#### Doc Al

Mentor
SpeeDFX said:
Thanks, that helped a lot. I now understand 2 of the 3 equations. I worked out the 2nd one and the 3rd one by myself, but now I'm a bit confused as to how the acceleration "a" of the 2 masses connected by the string is related to the acceleration "A" of the entire cart M.
The first thing to realize is that "a" is the acceleration of m1 with respect to the cart. But the cart itself is accelerating to the left with an acceleration of "A" with respect to the ground. When you apply Newton's 2nd law to m1, you must use the acceleration of m1 with respect to the ground, which is "a - A". (Newton's law cannot be applied, without correction, from an accelerating reference frame.)

This is my thought process...
The tension T acts on all 3 objects, m1, m2, and M. If I just look at the the horizontal accelerations, then the tension T causes m1 to accelerate to the right, and also causes M to accelerate to the left.
So far, exactly right. The horizontal force on m1 is T to the right; The horizontal force on M is T to the left.
The magnitude of both the force T onto m1 and M is kindof...split.. so that the the magnitude of the force required to move m1 PLUS the magnitude of the force required to move M is T.
The tension is not "split". It pulls left on M and right on m1. (Since the rope is massless, this is equivalent to M and m1 pulling on each other. The rope merely transmits the force between them. Think Newton's 3rd law.)
That makes me think of the following equation...

(m1)(a) + (M)(A) - T = 0, but I think that's wrong.
Yes it's wrong. It looks like you haven't decided which mass to apply Newton's law to! Analyze one mass at a time, please.
According to the equation (1) in the solution, I'm supposed to relate m1 with A, but I'm not able to grasp this relation.
It all hinges on expressing the acceleration of m1 with respect to the ground (which is an inertial frame of reference) in terms of "a" and "A". See my comments above.

Sorry for writing so much on a seemingly simple problem. I dont know why this one is so difficult for me.
I remember struggling for weeks on certain problems that I now think trivial, so I know the feeling.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving