# Pulley problem with two masses

## Homework Statement [/B]

F = m*a

## The Attempt at a Solution

[/B]
Hello, i think i understand d and e so only the first three tasks are important.

a) a = (m2 - m1)/(m1 + m2) *g -> (5kg - 2kg)/(2kg+5kg) *10ms^-2 = 4.29 ms^-2

T = m1 (a+g) -> 2kg (4.29ms^-2 + 10ms^-2) = 28 N

So i know the answer is 20N but im not sure why i got it wrong. Is it because block B stands on the ground and because of that the acceleration is 0?

b) The upward force should be the same magnitude as the downward force on block B.

Downward force = weight = 5kg * 10ms^-2 = 50N

Upward force = only normal force? = 50N

My problem here lies in my mistake from a i think. So does block B has tension force on it? Im kinda confused because block B rests on ground, so does it even have tension then? I mean if the acceleration from block A is really 0, then shouldnt the tension be zero as well?

c) total downward force of the pulley = weight from block A and B + weight of the pulley

20N + 50N + 15N = 85N

These should be the only forces acting downwards on the system but apparently this is wrong too. It should be 55N but i´m not sure why it should be this value. The tension force should be pointing upwards and the weight downwards, even if i subtract the tension force i wont get 55N.

Sorry for so much text. I hope someone can clarify this to me. Thanks for every answer.

#### Attachments

• 218.6 KB Views: 712

Related Introductory Physics Homework Help News on Phys.org
Orodruin
Staff Emeritus
Homework Helper
Gold Member
a) a = (m2 - m1)/(m1 + m2) *g -> (5kg - 2kg)/(2kg+5kg) *10ms^-2 = 4.29 ms^-2
Nothing is accelerating. B is supported by the ground below ...

• Krashy
All right thanks, whats what i thought so a should be clear, but then is there tension force on block B?

CWatters
Homework Helper
Gold Member
It's a statics problem so.
For a) Draw a free body diagram for block A. What two forces act on block A? What do they sum to?
For b) Draw a free body diagram for block B.

• Krashy
I drew one for both blocks. For block a i have the weight downwards = 20N and now i know the tension upwards = 20N , so theres no resultant force on A.
For block b i have the weight downwards = 50N and the normal force upwards which should also be = 50N i think

CWatters
Homework Helper
Gold Member
I drew one for both blocks. For block a i have the weight downwards = 20N and now i know the tension upwards = 20N , so theres no resultant force on A.
Correct.

For block b i have the weight downwards = 50N and the normal force upwards which should also be = 50N i think
Yes weight is 50N but there is another force on B in addition to the weight and normal force.

• Krashy
I see, i guess its tension force = 20N. So the normal force should be 50N - 20N = 30N
all right thanks i thought because block B is touching the ground there wouldnt be any tension.

• CWatters
Orodruin
Staff Emeritus
Homework Helper
Gold Member
I see, i guess its tension force = 20N. So the normal force should be 50N - 20N = 30N
all right thanks i thought because block B is touching the ground there wouldnt be any tension.
The tension in the rope cannot change from one point to the other (in the absence of other forces in the direction of the rope).

• Krashy
Ah all right thanks for the information now i know how to calculate a and b, but why is c wrong? I dont really get why the total downward force takes something else into account

• Abu
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Did you try drawing a FBD for the pulley?

What are the individual contributions based on what we just said?

• Krashy
i think there are 3 forces acting on the pulley, the weight of the pulley acting downwards and the two tension forces acting upwards.
So the only force thats acting downward from the pulley should be just its weight, right?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
How are the tension forces acting upward? What are the third law pairs of the tension forces?

• Krashy
all right i think i understand now, One is the tension acting from the pulley on the masses and the other is the tension acting from the masses on the pulley and because they are a third law pair they are identical in size and act in the opposite direction. So the total downward force is the 15N weight from the pulley + 2 times the 20N tension forces of the pulley on the masses, so the total downward force is 15N + 20N + 20N = 55N
Thanks for the help!

• CWatters and Delta2