# Thread:

1. ## Physics Question

I just had my last physics lesson before my exam, but I have one question where I disagree with my teacher.

A dynamics Trolley A of mass 5kg is placed on a horizontal board. It is connected to block B (a hanging mass) of mass 2kg by a light, inextensible string over a frictionless pulley.

Assuming no friction force on the trolley, calculate the magnitude of the acceleration of the trolley.
My teacher said use F = ma where F = 2(9.8) and m = 5kg. Which yields the answer 3.92 m/s²

I decided that the acceleration on the trolley and the hanging mass should be the same. So:

Trolley: $a = \frac{T}{M_a}$ (1)where $M_a$is the mass of the trolley, T is tension

Hanging mass: $a = \frac {F}{M_b}$where $M_b$ is the mass of the hanging mass.

$a = \frac{M_b g- T}{M_b}$ where g is acceleration due to gravity.
= $g -\frac{T}{M_b}$ (2)

Equating equations (1) and (2), I get 2.8 m/s². I'm also aware that using the formula F = ma where m = 7, F = 2g also gives me this answer, but I don't know why. Is my teacher wrong, or am I just over complicating things? Can someone clarify this problem for me?

2. hi there,

both methods are correct just that the way you do it is actually the longer way

what your teacher did was he/she solves the hanging mass first in order to obtain the tension in the string which then allows the usage of Newtons second law.

What you did was, you are solving both the trolley and the hanging mass at the same time

3. seems to me your making a mistake

you have for the trolley:

acceleration = tension / mass

and for the hanging mass

acceleration = force / mass and you rearrange this to be force = mass x gravity

so for the hanging mass you have force = 2kg x -9.8 m/s^2

but remember that the force acting on the hanging mass is the same as the force of the tension in the rope for the trolley

so then you should get

acceleration of trolley = force of hanging mass / mass of trolley

look over your equations im not sure how you got the one that reads

acceleration of trolley = mass hanging x gravity - tension / mass hanging

using the first two equations i dont see how to get to that one maybe you can show me

4. if the accelerations are the same then the masses have to be the same

try the experiment using a 10 kg trolley and a 0.1 kg hanging mass...
the hanging mass will always have acceleration of 9.8 m/s^2 down but do you think the 10 kg trolly will?

5. Originally Posted by finch41
if the accelerations are the same then the masses have to be the same

try the experiment using a 10 kg trolley and a 0.1 kg hanging mass...
the hanging mass will always have acceleration of 9.8 m/s^2 down but do you think the 10 kg trolly will?
Sorry but you're totally wrong. The acceleration of the hanging mass is defintely NOT g under these conditions. Unless of course the tension in the string is zero .........

6. sorry

7. Originally Posted by mr fantastic
Sorry but you're totally wrong. The acceleration of the hanging mass is defintely NOT g under these conditions. Unless of course the tension in the string is zero .........
what if friction forces were zero?

8. Originally Posted by finch41
what if friction forces were zero?
No, not even if friction forces were zero.

The fact is that the net force on the hanging mass m is (taking downwards as the positive direction) mg - T. Then:

$ma = mg - T \Rightarrow a = \frac{mg - T}{m}$.

a = g only if T = 0. If T = 0 then the string is slack .....

I suggest you thoroughly review (learn?) the application of Newtonian mechanics to this sort of problem ......
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