A Plane, A pulley and two boxes

In summary, the conversation discusses a physics problem involving a plane and friction. The question asks for the minimum and maximum values of m1 to prevent acceleration of the system. The conversation includes a diagram and equations to solve for these values, with the maximum value being significantly larger than the minimum value. The conversation also raises the possibility of additional factors that may affect the maximum value.
  • #1
Divergent13
48
0
Hi everyone here is the picture of the this plane/friction problem.

http://phasedma.com/uploaded/Physics problem.JPG

The question asks what are the minimum and maximum values of m1 in the figure to keep the system from accelerating...

Take µk = µs = 0.50
------------------------


Ok when drawing my free body diagrams I have come up with this method of solving the problem... tell me if you agree.

The force of friction can either be up or down the slope, if m2 = 0 or sufficiently small, then m1 would tend to slide down the plane. so Ffr would be directed Up the incline.

We know that Newtons second law for the y direction (i chose my xy coordinate axes along the plane-- IE horizontal x being the plane) shows that

Fnormal - m1*g*cos(30) = m1*ay = 0

since there's no y motion Fnormal = m1*g*cos(30)

Now for the x motion... For the first case (smallest m1) f = ma shows that

m1*g*sin(30) - Ftension - Ffr = m1*ax <---- x direction

since we want ax to be 0, we can solve Ftension since that's related to m2.

Since Ffr can be AT MOST µs * Fnormal= µs*m1*g*cos 30 the minumum value m2 can have to prevent motion (ax = 0) is (after dividing by g)

m2 = m1 * sin(30) - µs*m1*cos(theta).

And then finding the max value wouldn't be much more difficult from there since we already set up our equations..

Am I correct here? If you are willing can someone work it --- what range do you get for the mass?

Thanks for you help. Mechanics gets soooo tough!
 
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  • #2
Divergent13 said:
The question asks what are the minimum and maximum values of m1 in the figure to keep the system from accelerating...
Some of your thinking seems OK, but you are getting a little mixed up. First of all, your diagram shows m2 (the hanging mass) as being fixed at 5 kg, so I don't know why you are solving for m2!

The smallest value of m1 would just prevent it from sliding up the plane. Taking up the plane as positive, the forces on m1 (I just call it m) are:
-mg sin(30) -μmg cos(30) + T = ma = 0
(note that T must equal 5g)
You can solve this for the minimum value of m.

The maximum value of m1 would just prevent it from sliding down the plane. The forces on m in this case are:
-mg sin(30) +μmg cos(30) + T = ma = 0
You can solve this for the maximum value of m.

The difference between the two cases, as I'm sure you realize, is that the friction acts in different directions.
 
  • #3
Interesting, I see why the tension is 5g, and I get a minimum value of 5.35 kg which seems appropriate... but my max value is nearly 75kg!

Is that correct? Or is there something else that we should consider for max value...
 
  • #4
I think that's correct. Know that 75 kg also causes more friction.
 
  • #5
Thanks urban... but wow-- that is a huge difference.
 

Q1: How does a pulley work in this setup?

A pulley is a simple machine that uses a grooved wheel and a rope or belt to change the direction of a force. In this setup, the pulley is used to lift the boxes off the ground by pulling down on one side of the rope while the other side is attached to the boxes.

Q2: What are the forces acting on the boxes?

There are two main forces acting on the boxes in this setup: the force of gravity pulling the boxes down towards the ground, and the tension force from the rope pulling the boxes up. The force of gravity is greater, which is why the boxes are initially at rest on the ground.

Q3: How does the weight of the boxes affect the system?

The weight of the boxes is a crucial factor in this setup. The heavier the boxes, the greater the force of gravity pulling them down, and the more force is required to lift them with the pulley system. This can be demonstrated by adding more weight to the boxes and observing how it affects the system.

Q4: Can the direction of the pulley affect the system?

Yes, the direction of the pulley can affect the system. If the pulley is facing downward, the force of gravity will be acting against the direction of the tension force from the rope, making it more difficult to lift the boxes. If the pulley is facing upward, the force of gravity will be in the same direction as the tension force, making it easier to lift the boxes.

Q5: What is the purpose of the plane in this setup?

The plane serves as a flat surface for the boxes to slide on. This allows the boxes to move more easily and smoothly as they are being lifted by the pulley. The plane also prevents the boxes from falling off the edge of the table as they are being lifted.

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