Pulley system with 100 N force and two masses

[Mark Rice]

Homework Statement

The attached picture is essentially a little question a teacher gave me during revision today. It is not compulsary, but it is really irritating me not knowing how to do it. I'm sure it was made to be intentially hard as I was the only person given the problem. Can anyone shed light on this proof? The picture shows a pully style system, with an upward force of 100 N. I need to prove the acceleration of the 2 kg mass and the 5 kg mass (0.2 m/s/s and 15.2 m/s/s respectivly)

Homework Equations

Obviously it is worked out with F(unbalanced) = mass * Acceleration. And working backwords knowing the accelerations, i know that the unbalanced force to produce the required acceleration on each mass is 1 N for the 5 kg mass and 30.4 N for the 2kg mass. (Gravitational field strength = 9.8 m/s/s)

The Attempt at a Solution

No idea at all, spent a double period (100 minutes) trying to work it out. I'm clearly either missing something incerdibly obvious and simple, or this question is outwith my knowledge capabilites?

 

[LawrenceC]

Hint:

Write an equation that relates all the forces up (100 N) to all the forces down. Forces down include items such as weights and Newton Second Law forces.

What you say about the tension on either side of the pulley?

 

[tms]

Forget for the moment the answers you are given; just look at the physics. Start by looking at the forces on each block and on the pulley.

 

[Mark Rice]

Hint:

Write an equation that relates all the forces up (100 N) to all the forces down. Forces down include items such as weights and Newton Second Law forces.

What you say about the tension on either side of the pulley?

Well the total Unbalanced force of the system would be 31.4N i think (100-((2+5) * 9.8) =31.4 N and the tension on the right side would be 49 N and the left side would have the same tension ?

 

[Mark Rice]

Forget for the moment the answers you are given; just look at the physics. Start by looking at the forces on each block and on the pulley.

On each block there is the weight down and tension up ?

 

[Mark Rice]

PS - someone stated in my class that if i just 'halfed' the 100N and made it 50N acting upwards on each block, that would give the required unbalanced forces (ie 50 up, 19.6 down Fun= 30.4 which gives the 15.2 m/s/s). Is this just a coincidence with wrong physics, or the answer for some reason ?

 

[tms]

On each block there is the weight down and tension up ?

Yes, and on the pulley ...

 

[LawrenceC]

The tension on the left equals the tension on the right. So it's 50 N on each. What someone stated is indeed correct and not a coincidence with wrong physics.

If you equate the two tensions you'll have one equation with two unknown accelerations. The other equation comes from the summation of all forces in the whole system. It will also contain both accelerations as unknowns. So all you have to do is solve them simultaneously.

Either method is correct.

 

[Mark Rice]

The tension on the left equals the tension on the right. So it's 50 N on each. What someone stated is indeed correct and not a coincidence with wrong physics.

If you equate the two tensions you'll have one equation with two unknown accelerations. The other equation comes from the summation of all forces in the whole system. It will also contain both accelerations as unknowns. So all you have to do is solve them simultaneously.

Either method is correct.

Ahhhh, that's been annoying me all night, i think i kind of get it, thanks very much :)