- #1

- 5

- 0

the pulley is massless and frictionless.

What I have to do is determine the acceleration of the system and the tension force in the rope. How would I find the acceleration?

You should upgrade or use an alternative browser.

- Thread starter physicsnub
- Start date

- #1

- 5

- 0

the pulley is massless and frictionless.

What I have to do is determine the acceleration of the system and the tension force in the rope. How would I find the acceleration?

- #2

- 85

- 0

Can u find the accln now?

Sridhar

- #3

Doc Al

Mentor

- 45,093

- 1,398

- #4

- 72

- 0

Just use the equations

m

and

T - m

To find the net force acting on each.

- #5

- 5

- 0

Would the Acceleration be 12m/s^2, assuming gravity is 10m/s^2?

I have not yet been introduced to those specific equations, I have just kind of been trying to figure this out on my own, could you explain them a little bit?

Originally posted by AD

Just use the equations

m_{1}g - T = m_{1}a

and

T - m_{2}g = m_{2}a

To find the net force acting on each.

I have not yet been introduced to those specific equations, I have just kind of been trying to figure this out on my own, could you explain them a little bit?

Last edited:

- #6

- 72

- 0

No. Think about it. How could the acceleration be greater than g?

- #7

- 5

- 0

- #8

- 85

- 0

No the accln is not and will not be greater than g:

This is because, the weight of both the bodies will act downwards while the accln on the lighte body due to the heavier body will act upwards. (Get the picture???) Hence the force eqtn wil be like this:

[tex]m_{1} * g - T = m_{2}*(g-a)[/tex]

So that the net accln of the lighter mass will not be greater than g.

Sridhar

This is because, the weight of both the bodies will act downwards while the accln on the lighte body due to the heavier body will act upwards. (Get the picture???) Hence the force eqtn wil be like this:

[tex]m_{1} * g - T = m_{2}*(g-a)[/tex]

So that the net accln of the lighter mass will not be greater than g.

Sridhar

Last edited:

- #9

Doc Al

Mentor

- 45,093

- 1,398

No. Do you understand the equations that AD laid out for you?Originally posted by physicsnub

but it has a 1.2 kg mass slowing it down a bit, so would i just take 10 - 1.2 kg for the acceleration?

You are correct that if the mass were not connected to the rope, it would fall with acceleration = g. But it is connected to the rope, which pulls with tension T, thus reducing its acceleration. To find out what the acceleration is, don't cut corners... solve the equations!

- #10

- 72

- 0

m

The net force on the larger mass, m

The accleration of both masses is equal but the smaller one is travelling upward and the larger one is travelling downward.

For the second equation

T - m

The net force on the lighter mass is equal to the tension in the rope minus its weight, since the tension is greater than the weight because it is moving upwards.

What you should do is add these two equations and solve for 'a.'

- #11

- 72

- 0

?

Haven't you already posted that?

Haven't you already posted that?

- #12

- 5

- 0

The first equation m1g - T = m1a

for T do i just do T = (3.2kg)*(10m/s^2) ?

Would the first equation look like (3.2kg)*(10m/s^2) - 32 = (3.2kg)a ?

- #13

Doc Al

Mentor

- 45,093

- 1,398

Go up to AD's last post in this thread. Read itOriginally posted by physicsnub

The first equation m1g - T = m1a

for T do i just do T = (3.2kg)*(10m/s^2) ?

Would the first equation look like (3.2kg)*(10m/s^2) - 32 = (3.2kg)a ?

- #14

- 5

- 0

well if thats not how i would go about starting that problem, i have no idea what hes telling me to do ive been trying to do it for a while now. i guess ill just wait until we get to this in class because this isnt really going anywhere, thanks though.

Last edited:

- #15

Sko

- #16

Doc Al

Mentor

- 45,093

- 1,398

- #17

Sko

net acceleration= -4.5 m/[tex]s^2[/tex]

tension=17 N

Share: