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Blocks on inclined plane

  • Thread starter techie86
  • Start date
  • #1
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Homework Statement


We have an inclined plane, 20 degrees, with two blocks on it. The higher block (m2) is 15 kg and the lower block (m1) is 20 kg. A 200N force is applied to the blocks and they move up the inclined plane at a constant speed. What force does the 15 kg block exert on the 20 kg block?


Homework Equations


F = ma


The Attempt at a Solution


My thinking is as follows. The net force in the parallel direction must be zero since we have a constant speed. Therefore, the sum of the forces from blocks 1 and 2 must equal 200N.

m1*g*sin(20) + m2*g*sin(20) = 200N

We want the force that the higher block (m2) exerts on the lower block (m1).

m2*g*sin(20) = 200N - m1*g*sin(20)... so we need to find 200N - m1*g*sin(20).

This comes out to: 200N - (20kg)(-9.8 m/s^2)*sin(20 deg) = 200 - -67 = 267 N.

Does this answer make sense to anyone?

Thanks!
 

Answers and Replies

  • #2
PeterO
Homework Helper
2,425
46

Homework Statement


We have an inclined plane, 20 degrees, with two blocks on it. The higher block (m2) is 15 kg and the lower block (m1) is 20 kg. A 200N force is applied to the blocks and they move up the inclined plane at a constant speed. What force does the 15 kg block exert on the 20 kg block?


Homework Equations


F = ma


The Attempt at a Solution


My thinking is as follows. The net force in the parallel direction must be zero since we have a constant speed. Therefore, the sum of the forces from blocks 1 and 2 must equal 200N.

m1*g*sin(20) + m2*g*sin(20) = 200N

We want the force that the higher block (m2) exerts on the lower block (m1).

m2*g*sin(20) = 200N - m1*g*sin(20)... so we need to find 200N - m1*g*sin(20).

This comes out to: 200N - (20kg)(-9.8 m/s^2)*sin(20 deg) = 200 - -67 = 267 N.

Does this answer make sense to anyone?

Thanks!
have you considered whether there is any friction, as I don't think you said the surface was smooth?
 
  • #3
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Oh dear, yes I forgot! The kinetic friction is 0.1. If I had to guess, this might change my answer as:

Fparallel - Ffriction for each block...

m1*g*sin(20)-0.1*m1*g*cos(20) + m2*g*sin(20)-0.1*m2*g*cos(20) = 200N

We want the force that the higher block (m2) exerts on the lower block (m1).

m2*g*sin(20) = 200N - m1*g*sin(20) + 0.1*m1*g*cos(20) + 0.1*m2*g*cos(20)... so we need to find 200N - m1*g*sin(20) + 0.1*m1*g*cos(20) + 0.1*m2*g*cos(20).

This comes out to: 200N - 20*-9.8*sin(20) + 0.1*20*-9.8*cos(20) + 0.1*15*-9.8*cos(20) = 234.8N

Maybe this is right?
 
  • #4
PeterO
Homework Helper
2,425
46
Oh dear, yes I forgot! The kinetic friction is 0.1. If I had to guess, this might change my answer as:

Fparallel - Ffriction for each block...

m1*g*sin(20)-0.1*m1*g*cos(20) + m2*g*sin(20)-0.1*m2*g*cos(20) = 200N

We want the force that the higher block (m2) exerts on the lower block (m1).

m2*g*sin(20) = 200N - m1*g*sin(20) + 0.1*m1*g*cos(20) + 0.1*m2*g*cos(20)... so we need to find 200N - m1*g*sin(20) + 0.1*m1*g*cos(20) + 0.1*m2*g*cos(20).

This comes out to: 200N - 20*-9.8*sin(20) + 0.1*20*-9.8*cos(20) + 0.1*15*-9.8*cos(20) = 234.8N

Maybe this is right?
I doubt it.

It takes only 196 N to LIFT a 20kg block. It is going to take a lot less than that to move it up the slope.

The force acting down the slope on the 20kg block are a component of the weight force - less than half the 196 N weight force, plus the friction force, less than 0.1 x the weight force, so the answer will be less than 0.6 x 196 N.
By correctly applying the sine and cosine factors you should get the real answer.

btw: When a 200N force is applied to this pair of masses I don't think it will move at constant speed if μk = 0.1. It will accelerate up the slope.
 

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