What Mistake Was Made in Finding the Speed of the Boxes?

[LokLe]

Homework Statement

In the figure, two boxes, each of mass 24 kg, are at rest and connected as shown. The coefficient of kinetic friction between the inclined surface and the box is 0.16. Find the speed of the boxes just after they have moved 4.3 m. (in m/s) g=9.8ms-2

Relevant Equations

Kinematic equations
Conservation of mechanical energy
F = ma

Hi. I cannot find the correct answer to the problem.

Attempt:
Net Fx = T - Fg - fk = 0
Fk = 0.16(mgcos30) = ma
a = 1.3579 m/s^2

v^2 = u^2 + 2as
v^2 = 0 + 2*1.36*4.3
v = 3.417 m/s

Answer: v = 3.9027 m/s

What did I do wrong here?

 

[PeroK]

I'm not sure I understand what you've done. Have you taken the kinetic friction as the accelerating force?

 

[LokLe]

Since F= ma, shouldn't kinetic friction force also be equal to ma? I used the acceleration from Fk = ma to calculate the speed of the movement of the box.

 

[PeroK]

Since F= ma, shouldn't kinetic friction force also be equal to ma? I used the acceleration from Fk = ma to calculate the speed of the movement of the box.

What happens if the ramp is frictionless? What happens if $F_k$ is huge? Huge friction equates to huge acceleration?

 

[LokLe]

Use Fg (force of gravity) to calculate the acceleration?

 

[PeroK]

Use Fg (force of gravity) to calculate the acceleration?

There are some notes on your diagram which suggests that someone has tried to solve the problem using a free-body diagram and a calculation of the net force. Was that not you?

That's the right approach, by the way!

 

[LokLe]

That was me. But I do not know which method I should use to solve it.
Edit: In the notes, I found that after calculation, the net force in x and the net force in y are not the same.

 

[PeroK]

That was me. But I do not know which method I should use to solve it.
Edit: In the notes, I found that after calculation, the net force in x and the net force in y are not the same.

Have you ever done a problem anything like this before?

You can use forces, or you can use energy.

 

[LokLe]

Are there forces that I had left out?
Also, I actually did a problem similar to this before, but I cannot find the answer to that problem as well.

 

[PeroK]

Are there forces that I had left out?
Also, I actually did a problem similar to this before, but I cannot find the answer to that problem as well.

We're not supposed to give you the solution, so you are expected to have some idea. You seem to have all the forces: gravity, friction, tension and the normal force.

One thing you may be missing is that the magnitude of the acceleration of both blocks must be the same. Does that help?

 

[LokLe]

Do you mean that I should add ma to the Net Fx so that both sides would have the acceleration component?

 

[PeroK]

Do you mean that I should add ma to the Net Fx so that both sides would have the acceleration component?

I mean you could analyse the forces on each block with the constraint that the acceleration must have the same magnitude in each case. That might help you find $T$.

 

[haruspex]

Net Fx = T - Fg - fk = 0

shouldn't kinetic friction force also be equal to ma?

This seems to be a basic misunderstanding.
The general equation is net force = mass x acceleration. The equation you have written assumes the acceleration is zero. Friction is just one of the forces contributing to the net force. It has no special relationship to the acceleration.

 

[LokLe]

Sorry I am still not able to solve the problem. I would be grateful if there are some hints available. Is the tension in x the same in y and since I do not know the tension force, how should I solve for the net work of the system?

 

[haruspex]

Sorry I am still not able to solve the problem. I would be grateful if there are some hints available. Is the tension in x the same in y and since I do not know the tension force, how should I solve for the net work of the system?

Let the tension be T, the normal force from the slope be N and the acceleration be a.
Write equations of the form $\Sigma F=ma$ for:
- the vertical direction of the hanging mass
- the upslope direction of the sliding mass
- the normal to the slope of the sliding mass
Three equations, three unknowns. Solve.

 

[LokLe]

Force Normal:
Fn = mgcos30 = 203.7N

Net Fx = (1/m)(Fg - fk - T) = a

Net Fy = (1/m)(T - mg) = a

Since magnitude of a in Net Fx and Net Fy are the same:
Fg - fk - T = T - mg
2T = mgsin30 - mgcos30*0.16 + mg
2T = 320
T = 160N

Fg - fk - T = ma
85 - 160 = 24a
a = -3.125 m/s^2 (?)

v^2 = u^2 + 2*(-3.129)*4.3
...

I am really sorry that I cannot calculate the correct answer. I believe there must be some mistakes in the equations or in my calculations.

 

[haruspex]

a = -3.125 m/s^2 (?)

Think about what this is telling you. Did you make a wrong assumption somewhere?

 

[LokLe]

I think I messed up the positive and negative signs somewhere. But the question does not tell me which direction the box is going so I can assume the direction of the boxes myself?

 

[LokLe]

I just found the answer.

Assuming the block is going towards the right, we can treat it as it is on the same line.

a = Net F/m total
a = (mg-mgsin30-mgcos30(0.16)) / (m+m)
a = 85/48
a = 1.771 m/s^2

v^2 = u^2 + 2as
v^2 = 0 + 2*1.771*4.3
v = 3.9027 m/s
So speed is 3.9027 m/s

Thanks for helping me solve the problem (It has bothered me for an entire week).

 

[haruspex]

I think I messed up the positive and negative signs somewhere. But the question does not tell me which direction the box is going so I can assume the direction of the boxes myself?

The correct approach is to consider what would happen without friction. Friction will act to oppose that relative motion of the surfaces in contact.

 

[LokLe]

If there is no friction, Net Fx = T - mgsin30 = ma
When there is friction Net Fx = T - mgsin30 - mgcos30*0.16 = ma

Net Fy = T - mg = -ma
T = mg - ma

mg - ma - mgsin30 - mgcos30*0.16 = ma
2ma = mg -mgsin30 - mgcos30*0.16
a = (mg -mgsin30 - mgcos30*0.16) / (2m)

I understand what you mean now. Thank you.