Force with friction problem, Please help

[AntonioMD]

Hey everyone, perhaps you can help me here. I'm having an argument with my classmates on this problem and I'm 99% certain I'm correct. Please provide any feedback.

Homework Statement

Block A is 3kg and is sitting on a horizontal surface, it is attached to a string of negligible mass to block B of 1kg, which is sitting on an inclined plane of 30 degrees to the right of Block A, over a pulley of negligible mass. Block A has a force of 2.3 N being applied to it in the positive direction of x. Both blocks have a coefficient of kinetic friction of .20, Solve for the acceleration of the system and the tension in the string.

Homework Equations

m1 = 3Kg
m2=1kg
coef. of kinetic friction: .20
Fap= 2.3N

Force of kinetic friction=Fn x Coefficient

The Attempt at a Solution

Applying Newtons second law to mass 1

T+Fapplied- force of kinetic friction=m1a

Applying Newtons second law to mass 2

m2gsin∅-T-force of kinetic friction on block 2= m2a
I solve for tension on both equations and then combine them to solve for acceleration:

T=m1a-Fapplied+Fk1

T=m2gsin∅-Fk2-m2a

a=(m2gsin∅-Fk2-Fk1+Fapplied)/(m1+m2)

This yields -.37N/4Kg

My fellow students claim that they're obtaining a positive acceleration but I don't understand how that's even possible.

You have a 3kg Block A with a kinetic friction of 3kg*9.8m/s^2 = 29.4N x .20 = 5.88N and an applied force of only 2.3N, meaning you're going to need a force greater than 3.58 N to keep Block A moving.

Block b has a kinetic friction of 1kg*9.8cos∅ = 8.5N x .20 = 1.7N and the only force acting on block B is gravity, with magniture m2gsin∅ = 1kg*9.8sin∅ = 4.5 N, so the net force acting on the block is only 2.8N. Does this not mean that it is impossible for block A to move along the positive direction of X? My acceleration is yielding negative because the kinetic friction force is greater than the net forces acting in the positive direction of X.

Can someone please confirm that I am correct or please explain to me why I am wrong. It would be greatly appreciated!

 

[PeterO]

Hey everyone, perhaps you can help me here. I'm having an argument with my classmates on this problem and I'm 99% certain I'm correct. Please provide any feedback.

Homework Statement

Block A is 3kg and is sitting on a horizontal surface, it is attached to a string of negligible mass to block B of 1kg, which is sitting on an inclined plane of 30 degrees to the right of Block A, over a pulley of negligible mass. Block A has a force of 2.3 N being applied to it in the positive direction of x. Both blocks have a coefficient of kinetic friction of .20, Solve for the acceleration of the system and the tension in the string.

Homework Equations

m1 = 3Kg
m2=1kg
coef. of kinetic friction: .20
Fap= 2.3N

Force of kinetic friction=Fn x Coefficient

The Attempt at a Solution

Applying Newtons second law to mass 1

T+Fapplied- force of kinetic friction=m1a

Applying Newtons second law to mass 2

m2gsin∅-T-force of kinetic friction on block 2= m2a
I solve for tension on both equations and then combine them to solve for acceleration:

T=m1a-Fapplied+Fk1

T=m2gsin∅-Fk2-m2a

a=(m2gsin∅-Fk2-Fk1+Fapplied)/(m1+m2)

This yields -.37N/4Kg

My fellow students claim that they're obtaining a positive acceleration but I don't understand how that's even possible.

You have a 3kg Block A with a kinetic friction of 3kg*9.8m/s^2 = 29.4N x .20 = 5.88N and an applied force of only 2.3N, meaning you're going to need a force greater than 3.58 N to keep Block A moving.

Block b has a kinetic friction of 1kg*9.8cos∅ = 8.5N x .20 = 1.7N and the only force acting on block B is gravity, with magniture m2gsin∅ = 1kg*9.8sin∅ = 4.5 N, so the net force acting on the block is only 2.8N. Does this not mean that it is impossible for block A to move along the positive direction of X? My acceleration is yielding negative because the kinetic friction force is greater than the net forces acting in the positive direction of X.

Can someone please confirm that I am correct or please explain to me why I am wrong. It would be greatly appreciated!

IF the acceleration was negative, the friction forces would be in the opposite direction!

Perhaps these masses don't go in either direction?

 

[Doc Al]

Double check that you have the problem statement completely correct. Was a diagram included?

 

[AntonioMD]

There was a diagram included but the problem is exactly as stated. I've attached an image of the problem.

 

[Doc Al]

Double check that you haven't reversed m1 and m2.

 

[PeterO]

There was a diagram included but the problem is exactly as stated. I've attached an image of the problem.

Just a hint when solving this sort of problem ...

Instead of randomly assigning m₁ and m₂ to the masses, call the 1 kg mass m₁ and the 3 kg mass m₃. That way you are never in doubt about which mass you are dealing with.
And yes, I do have problems that include three masses m₅, m₇ and m₁₂.

Note: I am not saying you have got the masses back to front here!