How Much Force Does a 15 kg Block Exert on a 20 kg Block on an Inclined Plane?

[techie86]

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!

 

[PeterO]

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?

 

[techie86]

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?

 

[PeterO]

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 μ_k = 0.1. It will accelerate up the slope.

 

[asteriatic]

Yes, your answer makes sense. In this scenario, the force that the 15 kg block exerts on the 20 kg block is 267 N. This is because the 200N force is being split between the two blocks, with a larger portion being applied to the 20 kg block due to its greater mass. This demonstrates the concept of Newton's Third Law, where for every action there is an equal and opposite reaction. The 15 kg block exerts a force on the 20 kg block in the opposite direction, allowing both blocks to move up the inclined plane at a constant speed.