Blocks on inclined plane

  • Thread starter techie86
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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
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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!
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
47
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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