Newton's Third Law and Pulley system

[StephenDoty]

The lower block in the figure is pulled on by a rope with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.30. The coefficient of kinetic friction between the lower block and the upper block is also 0.30. What is the acceleration of the 2.0 kg block??

Forces of A: T-(.3*1kg*9.8m/s/s)= (1kg)a
Forces on B: 20N - T- (.3 * 3kg * 9.8m/s/s)= (2kg)a

If you Add them together: 20N-(.9*9.8)-(.3*9.8) = 3a
20N-11.76 = 3a
8.24/3 = a = 2.74667m/s/s

Should it be 3kg as the mass used in finding the kinetic friction force in the formula for the forces on B or should it be 2kg? And should it be 2kg as the mass on the right side of the equation for the forces on B or should it be 3kg?

the answer of 2.75 is not correct. What am I doing wrong?

Thank you.

Stephen Doty

 

[kdv]

The lower block in the figure is pulled on by a rope with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.30. The coefficient of kinetic friction between the lower block and the upper block is also 0.30. What is the acceleration of the 2.0 kg block??

Forces of A: T-(.3*1kg*9.8m/s/s)= (1kg)a
Forces on B: 20N - T- (.3 * 3kg * 9.8m/s/s)= (2kg)a

If you Add them together: 20N-(.9*9.8)-(.3*9.8) = 3a
20N-11.76 = 3a
8.24/3 = a = 2.74667m/s/s

Should it be 3kg as the mass used in finding the kinetic friction force in the formula for the forces on B or should it be 2kg? And should it be 2kg as the mass on the right side of the equation for the forces on B or should it be 3kg?

the answer of 2.75 is not correct. What am I doing wrong?

Thank you.

Stephen Doty

We can't see the picture yet. But it seems as if you are forgetting one force on the bottom block. If there is friction on the top block one wway, there must be the corresponding reaction force on th ebottom block acting the opposite way. So there are TWO friction forces on the bottom block. On friction due to the ground and one due to the top block

 

[StephenDoty]

so do you add those two frictional forces
or do you subtract them since they are equal and opposite?

If you add them then would you please explain why you would not subtract them since they are equal and opposite?

Thank you

 

[kdv]

so do you add those two frictional forces
or do you subtract them since they are equal and opposite?

Be careful, forces that are action-reaction pairs NEVER ACT ON THE SAME OBJECT.

So the two frictions forces on the lower block are NOT part of an action reaction pair. One is due to the ground, the other is due to the top block.

Now, you have to figure out in what direction they are acting. The friction of th etop block on the bottom block will be opposite to the friction of the bottom block on the top block (these two do form a pair). I can't tell because I don't know if the friction force between the two blocks is static or kinetic (is the top block sliding against the bottom block or are they both accelerating as one object)? I need a better description of the situation

 

[StephenDoty]

as the bottom block slides to the right the top box is sliding to the left.

 

[Doc Al]

Forces of A: T-(.3*1kg*9.8m/s/s)= (1kg)a

Good.

Forces on B: 20N - T- (.3 * 3kg * 9.8m/s/s)= (2kg)a

You only counted the friction from the bottom surface. But the top block also exerts a friction force on the bottom block.

Should it be 3kg as the mass used in finding the kinetic friction force in the formula for the forces on B or should it be 2kg?

In finding the friction between any two surfaces, what counts is the normal force. Between the bottom block and the floor the normal force is the total weight of both masses = 3 kg. (But that's not the only friction force on the bottom block.)

And should it be 2kg as the mass on the right side of the equation for the forces on B or should it be 3kg?

The mass in F = ma is the mass of the object you are analyzing. For block B, it's the mass of block B = 2 kg.

 

[mancini0]

So they are connected by a rope, so the acceleration for both is the same. That is, they have the same "a" value. So do a F = ma for block one, and do a F = ma for block 2, and solve for
"a" in each equation. Then maybe you could set them equal to each other? I'm working on this problem now, I'll let you know if it works.

 

[Doc Al]

So they are connected by a rope, so the acceleration for both is the same. That is, they have the same "a" value.

Yes, they have the same magnitude of acceleration.

So do a F = ma for block one, and do a F = ma for block 2, and solve for
"a" in each equation. Then maybe you could set them equal to each other?

Yes, apply Newton's 2nd law to each block separately. Then you can solve them together to get the acceleration.

If you call the acceleration of the bottom block +a, what would be the acceleration of the top block?

 

[mancini0]

BINGO! It worked. So to contribute:
"The lower block in the figure is pulled on by a rope with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.30. The coefficient of kinetic friction between the lower block and the upper block is also 0.30.
Find the acceleration of the 2 kg block."

Equation for block 1 is F = M*A
The force for block 1 is Tension - (friction coefficent*mass*g)
the mass is 1 kg
the a is unknown, but we know that both blocks have the same acceleration, since they are connected.
So here is the final equation that describes the acceleration of block 1:
T-.3(1kg*9.8) =1kg*a
T-.3(9.8) = a (I just got rid of the 1kg, because 1 * anything is itself.)

now let's find an equation that describes the acceleration of block 2:
F = ma
20N - T -(.3*3kg*9.8 + .3(1kg*9.8) = 2kg*a
so:
a = {20N - T -(.3*3kg*9.8 + .3(1kg*9.8) } / 2

since a is the same for block a and b, set these two equations equal to each other, and solve for T.

this gives T = 4.706.

plug T back in one of the equations, and get a.
The first equation is easier to work with:
T - (.3*g) = a
4.706 -(2.94) = a
a = 1.766

this gave me the correct answer on mastering physics.