# Homework Help: Simple question:S help!

1. Sep 3, 2006

### Clef

doing the dynamics trolley exp. for newtons 2nd law. stuck on a couple of q's, i realise the concept but i am having trouble expressing it:)

1.why is the mass in net force=ma not just equal to the mass of the cart?

2.when calculating the force on the cart using mass times gravity, why isnt the mass of the cart included?

2. Sep 3, 2006

### Hootenanny

Staff Emeritus
I'm not familiar with this experiment, perhaps you could describe it.

3. Sep 3, 2006

### Clef

well, there's a pulley, with a mass (m1) attached to a string hanging off the edge of a table and a cart (m2) which is being dragged towards the edge of the table by this mass. it is quite a simple experiment and you just time how long it takes for the cart to travel a certain distace, whilst adding weights on top of it each trial to measure its acceleration. make sense?

4. Sep 3, 2006

### Hootenanny

Staff Emeritus
Perfect sense, something like this I imagine;
http://www.practicalphysics.org/imageLibrary/jpeg450/519.jpg [Broken]
Now, for your questions;
If there cart was empty you would be quite correct, m would equal the mass of the cart. However, since there are additional masses at rest on the cart; these masses must be accelerating at the same rate as the cart. Does this make sense?
Why do you think that the mass of the cart should be included?

Last edited by a moderator: May 2, 2017
5. Sep 3, 2006

### Clef

thats exactly what i meant, and to the second one , is it because the force of the mass balance accelerating because of gravity is the only force acting upon the cart?

6. Sep 3, 2006

### Clef

oh and i know what you mean to the 1st question. but the question is actually asking why to get the net force for the system you need to include the mass of the cart AND the mass of the weight on the side of the pulley.

7. Sep 3, 2006

### Hootenanny

Staff Emeritus
Almost. If we ignore friction, there are three forces acting on the cart. There is the weight of the cart (m2g), this is opposed by the normal reaction force (provided by the table) which is equal in magnitude to the weight of the cart but opposite in direction. There is also the tension in the string (which we assume to be uniform). Now, since the normal reaction force is equal but opposite to the weight of the cart there is no net force and hence no acceleration in the vertical plane. However, in the horizontal plane the only force acting is the tension in the string, therefore an acceleration is produced. Do you follow?

I forgot to welcome you to Physics Forums before

Last edited: Sep 3, 2006
8. Sep 3, 2006

### Hootenanny

Staff Emeritus
Ahh, sorry I misunderstood. Quite simply both masses are accelerating.

9. Sep 3, 2006

### Clef

yes wow, very good explanation, and thankyou:D wow, i love it here already..

10. Sep 3, 2006

### Hootenanny

Staff Emeritus
My pleasure