Pulley on an Incline: Finding Mass & Acceleration

[mmoadi]

Homework Statement

Flat weights on a picture are connected by a light rope by a light pulley. How much at most (at least) does mass m2 have to be to balance the pulley, if mass m1 = 1 kg. With how much acceleration do the weights move, if the mass m2 of the weight is replaced with a twice heavier (lighter) weight? Ratio of friction between the weight with mass m1 and the surface is 0.3

Homework Equations

F1= m1g sinα μk
F2= m2g
F1 – F2= a(m1 + 2m2)

The Attempt at a Solution

m1= 1 kg
μk= 0.3
α= 30°
m2=?
a= ?

For finding the m2:
F1 = F2
m1g sinα μk = m2g
m2= 0.15 kg

For finding the acceleration, if m2 is double the weight:
F1 – F2= a(m1 + 2m2)
g(m1 sinα μk – 2m2) = a(m1 + 2m2)
a= 1.51 m/s²

For finding the acceleration, if m2 is half the weight:
F1 – F2= a(m1 + ½m2)
g(m1 sinα μk – ½m2) = a(m1 + ½m2)
a= 3.87 m/s²

 

[mmoadi]

Can someone please tell me if I calculated correctly? Thank you!

 

[gamer_x_]

I don't understand where you're getting your angles from, unless one of the masses is also on a ramp? Can you describe the geometry of the setup a little better?

 

[mmoadi]

I attached the picture of the pulley system. Hope it will make the problem clearer.
Thanks for helping!

 

[rl.bhat]

Your calculation of F1 is wrong.
On m1 two forces are acting. What are they? Draw FBD.

 

[mmoadi]

OK, I added the forces to the picture.

F1 is decomposed into F1a and F1b and because F1a and -f1a cancel each other I assumed there is only one force left working on m1, which is F1b (= m1g sinα μk).

Where did I go wrong?

 

[mmoadi]

After looking at the picture and thinking really, really hard, I remembered the tension of the rope. Does this too has to be considered when calculating F1?

 

[rl.bhat]

Forces acting on m1 are the component of weight and frictional force in the downward direction and tension in the upward direction.
What is the frictional force?

 

[mmoadi]

Fk= μkN
N= mg

 

[mmoadi]

So the sum of the forces working in y direction are 0, because they cancel each other out.
But the sum of forces working on x direction taking tension into the consideration would be
T- m1g sinα?

 

[mmoadi]

I think I lost you.
So, we have tension of the rope, which is pulling in the direction upwards. Does this mean that tension is connected with m2 and I can write T= m2g?

If my upper conclusions are correct, does the formula for F1 now rewrites to:
F1= T- m1g sinα= m2g -m1g sinα, and I can now from this formula calculate m2?

I'm desperate!

 

[rl.bhat]

Fk= μkN
N= mg

N is m1g*cosθ

Τ - m1*g*sinθ - μ*m1*g*cosθ = m1a ...(1)
For m2
m2*g - T = m2*a...(2)
From eq 1 and 2 find a and equate them. Then solve for T.
Then substitute T is either eq 1 or 2 to find a.

 

[mmoadi]

Hello! for the second equation are you sure that the tention is negative? because i read that it always should be positive! and could you explain what to do to solve the acceleration and tension, that would help a lot! thank you in advance!

If the tension is positive in the second equation i think the m2 should be .24 kg, right!

 

[rl.bhat]

Body accelerates in the direction of the net force.In the second case, if 2m2 is moving down, then 2m2*g must be greater than T. So net force = 2m2*g - T = 2m2*a.

 

[mmoadi]

OK, let's start from the beginning.

The first part of the problem is asking us to find weight of m2 that will make the system to be in equilibrium.

As you see I made some approaches to solve this part of the problem, but I obviously didn't succeed. Can you please give me some hint?

I think that having m2 known will make the rest of the problem much much easier to solve, because I think I understand your explanation how to merge the formulas and finding T and a.

Thank you for your time and patience!

 

[rl.bhat]

Τ - m1*g*sinθ - μ*m1*g*cosθ = m1a ...(1)
For m2
m2*g - T = m2*a...(2)
In the above equations, put a = 0 and solve for m2.

 

[mmoadi]

So by putting a=0 into the above equations I ended with equation:

m1*g*sinθ + μ*m1*g*cosθ = m2*g

and solved for m2= 0.76 kg.

I really hope I'm correct!

 

[mmoadi]

So, if I understood correctly, by adding the equations for m1 and m2 I should get the formula for acceleration which would be

a= (m2g - m1*g*sinθ - μ*m1*g*cosθ) / (m1 + m2)

and playnig with weight (once double the weight, once half the weight for m2) I should be able to get the correct results?

 

[mmoadi]

I did the the calculations and get this results:

for 2m2 (double the weight) the a= 2.96 m/s²
for 1/2 m2 (half the weight) the a= - 2.7 m/s²

Did I get it right?