Pulley problem involves Friction static and Kinetic problem

In summary: Now when you put the other information about a = 1.52 m/s^2 into [2], you get T = 28.465 N which leads to u = 0.405 for the kinetic coefficient. The difference between these is 0.479 - 0.405 = 0.074. So in summary, the static coefficient is 0.479 and the kinetic coefficient is 0.405, with a difference of 0.074.
  • #1
Leocardinal
2
0

Homework Statement



M1 has a mass of 6.51 kg. It is on a horizontal surface, connected by a light string to a hook. Mass M2 can be increased smoothly by adding masses little at a time. The pulley has a negligible mass and no friction.

When M2 is 3.12 kg it begins to accelerate downwards at a rate of 1.52 m/s2. Calculate mu static -mu kinetic for M1 on the surface.

I was also given this hint, First calculate ms (from knowing the force it takes to begin moving) and then mk (from the given acceleration) and subtract

Homework Equations



F= Ma


Mass2
Fnety = T-Mg


Mass1

Fnetx= T-Fstatic

Fs= Mus*Fn

Fk= Muk* FN

The Attempt at a Solution



Mass2
Fnety = T-Mg
-4.74N= T-30.576N
T= 25.83

Mass1

Fnetx= T-Fkinetic
9.89N= 25.83-15.934
Fk= Muk*Fn
15.9348/63.798= 25.83

Muk= .24976

Fnet=0
T-Fs=
25.83

Fs= Mus* FN

25.836= Mus * FN
25.836/63.798

Mus=.405

Mus-Muk= .155
 

Attachments

  • prob71_upmasspulley.gif
    prob71_upmasspulley.gif
    1.9 KB · Views: 594
Physics news on Phys.org
  • #2
There is something odd about this solution and I can't quite pin it down. I think you are supposed to write F = ma for each mass as your starting point. This would give you 2 equations:
[1] T - u*m1*g = m1*a and [2] m2*g - T = m2*a
where T is the tension in the string between the two masses.
When a = 0, you get T = u*m1*g from [1] and when you put it into [2] you have m2*g = u*m1*g which gives u = m2/m1 = 0.479 for the static coefficient.
 
  • #3
2





Based on the information provided, the pulley problem involves both static and kinetic friction. The problem requires us to calculate the coefficients of static and kinetic friction for M1 on the surface. To do this, we can use the given hint to first calculate the static friction (Fs) by using the formula Fs = Mus * FN, where FN is the normal force and Mus is the coefficient of static friction. We can find FN by using the formula Fnet = ma, where Fnet is the net force acting on M1 and a is the acceleration of M1. From the given information, we can calculate the net force to be 9.89N and the acceleration to be 1.52 m/s2. Therefore, FN = 9.89N/1.52 m/s2 = 6.51 kg. Plugging this value into the equation for Fs, we get Fs = Mus * 6.51 kg.

Next, we can use the given information about M2 to find the coefficient of kinetic friction (Muk). We know that when M2 is 3.12 kg, it begins to accelerate downwards at a rate of 1.52 m/s2. Using the formula Fnet = ma, we can find the net force to be 15.9348N. We also know that T = 25.83N, where T is the tension in the string. Therefore, using the formula Fnetx = T - Fk, we can find Fk to be 9.89N. Plugging this value into the equation for Fk, we get Fk = Muk * 6.51 kg. Solving for Muk, we get Muk = 0.24976.

Now, we can calculate the difference between Mus and Muk by subtracting them. Mus - Muk = 0.405 - 0.24976 = 0.1552. Therefore, the difference between the coefficients of static and kinetic friction for M1 on the surface is 0.1552. This means that the coefficient of kinetic friction is lower than the coefficient of static friction, which is typically the case in real-world situations.
 

1. What is a pulley problem involving friction static and kinetic?

A pulley problem involving friction static and kinetic is a physics problem that involves analyzing the forces and movements of a system of pulleys with friction present. Friction can either be static, where the objects are at rest, or kinetic, where the objects are in motion.

2. How do you solve a pulley problem involving friction static and kinetic?

To solve a pulley problem involving friction static and kinetic, you first need to identify all the forces acting on the system, including the tension in the ropes and the forces of friction. Then, you can set up equations using Newton's laws of motion and solve for the unknown variables.

3. What is the difference between static and kinetic friction in a pulley problem?

In a pulley problem, static friction refers to the force that prevents an object from moving when it is at rest, while kinetic friction refers to the force that opposes the motion of an object that is already in motion. The value of static friction is typically greater than the value of kinetic friction.

4. How does the presence of friction affect the solution of a pulley problem?

The presence of friction in a pulley problem can make the solution more complex, as it adds an additional force to consider in the equations. Friction can also affect the tension in the ropes and the motion of the objects involved, so it is important to accurately account for it in the problem.

5. Are there any tips for solving a pulley problem involving friction static and kinetic?

One tip for solving a pulley problem involving friction static and kinetic is to draw a free body diagram of the system, clearly labeling all the forces involved. It is also helpful to break the problem into smaller parts and solve them separately before combining the solutions. Additionally, it is important to carefully consider the direction and magnitude of each force when setting up the equations.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
28
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
15K
  • Introductory Physics Homework Help
Replies
1
Views
7K
  • Introductory Physics Homework Help
Replies
7
Views
3K
Back
Top