Two Masses Connected by Pulley; find acceleration

In summary, if the system is in motion, the hanging mass will move to the right and the block on the table will move to the left.
  • #1
cassienoelle
63
0

Homework Statement


Mass m1 = 11.9 kg is on a horizontal surface. Mass m2 = 8.80 kg hangs freely on a rope which is attached to the first mass. The coefficient of static friction between m1 and the horizontal surface is μs = 0.526, while the coefficient of kinetic friction is μk = 0.168.
(There is a picture with M1 on top of a ...lets say table..and the m2 box is on the side of the table with a pulley on the corner)
If the system is in motion with m1 moving to the left, then what will be the magnitude of the system's acceleration? Consider the pulley to be massless and frictionless.


Homework Equations


Extra notes from professor:
The frictional force will always oppose the direction of relative motion (or impending motion). If the block on the horizontal surface is moving to the right, then the frictional force is to the left. If the block on the horizontal surface is moving to the left, then the frictional force is to the right. However, you should see from a force diagram of the block on the horizontal plane, the tension in the string is to the right in both cases (tensions can only pull).


The Attempt at a Solution


I've tried different things by examples my professor has given me, however those included angles so I'm not sure how to do this one.
 
Physics news on Phys.org
  • #2
It appears that the hanging mass is on the right side of the table and moving up, and the block on the table is moving left, correct? Draw a free body diagram of each block, identify the forces acting (with the help of your professor's hints), and use Newton 2 to solve for the acceleration . You have to solve 2 equations with 2 unknowns.
 
  • #3
Newton 2 being?
SigmaF = ma for each?
 
  • #4
Yes. Newton's 2nd law states that...

[itex]\Sigma[/itex]F = ma
 
  • #5
Okay. So...
For M1: there is natural, gravity, and tension ?
For M2: there is natural, gravity, weight, and tension?
 
  • #6
M1 T - m1*g = m1*a
M2 T + m2*g = m2*a
?
 
  • #7
M1: Normal (upward), Weight (downward), Tension (rightward), Friction (rightward)
M2: Tension (upward), Weight (downward)

M1's equation

[itex]\Sigma[/itex]F = m[itex]_{1}[/itex]g - m[itex]_{1}[/itex]g - T[itex]_{1,2}[/itex] - [itex]\mu[/itex]m[itex]_{1}[/itex]g = m[itex]_{1}[/itex]a

M2's equation

[itex]\Sigma[/itex]F = -m[itex]_{2}[/itex]g + T[itex]_{2,1}[/itex]= m[itex]_{2}[/itex]a

Correct?
 
  • #8
Okay. And w = mg F = ma now what?
 
  • #9
Now, you add your 2 equations together...

[itex]\Sigma[/itex]F = -m[itex]_{2}[/itex]g - [itex]\mu[/itex]m[itex]_{1}[/itex]g = m[itex]_{1}[/itex]a + m[itex]_{2}[/itex]a

Solve for a.
 
  • #10
okay. i guess i just don't understand how to get the equations.
 
  • #11
If you want, I can try to explain how to get these equations. :smile:
 
  • #12
In order to come up with these equations, you have to first figure out what forces are acting upon the objects in question.

The four basic ones when dealing with pulleys are:
1. Weight (otherwise known as gravitational)
2. Normal (otherwise known as natural in some cases)
3. Tension
4. Friction.

As you already know, weight forces tend to be in the pointed downwards. Normal forces tend to be pointed upwards.

Friction forces and Tension forces vary based on the circumstances.

So, let's consider this problem.

We have a weight force pointing downward and a tension force pointing upward on M2.

On M1, we have a weight force pointing down, a tension force pointing to the right, a friction force pointing to the right, and a normal force pointing upwards.

Weight force = -mg
T force for M1 = T force for M2
Friction force = u(N)
Normal force = mg

Now, when we sum our forces for the M1 object...

[itex]\Sigma[/itex]F = ma

The sum is just all of the above forces added together and set equal to ma.

[itex]\Sigma[/itex]F = -mg + mg + (-T) - u(N) = ma

Now, T is negative because it is pointing opposite of the direction of motion, and friction is negative for the same reason.

Now, for M2...

We have a W force = -mg and a T force.

So, [itex]\Sigma[/itex]F = T - mg = ma

T is positive because it is pointing upward and W is negative because it is pointing downward.


And that's a specific way of how to get the equations for this problem. :smile: Hopefully it cleared things up a bit.
 

What is the concept behind two masses connected by a pulley?

The concept behind two masses connected by a pulley is that they are connected by a rope or string that passes over a pulley, allowing them to move in opposite directions. This set-up is often used to demonstrate the relationship between mass, force, and acceleration.

How do you calculate the acceleration of two masses connected by a pulley?

The acceleration of two masses connected by a pulley can be calculated using the formula a = (m1 - m2)g / (m1 + m2), where m1 and m2 are the masses of the objects and g is the acceleration due to gravity (9.8 m/s^2).

What factors can affect the acceleration of two masses connected by a pulley?

The acceleration of two masses connected by a pulley can be affected by the masses of the objects, the tension in the rope, and any external forces acting on the system. Friction and air resistance can also play a role in the acceleration.

Can the acceleration of two masses connected by a pulley be negative?

Yes, the acceleration of two masses connected by a pulley can be negative if the masses are moving in opposite directions and one of the masses has a greater value. This would indicate that the system is decelerating or moving in the opposite direction of the net force.

How does the angle of the pulley affect the acceleration of two masses?

The angle of the pulley can affect the acceleration of two masses connected by a pulley by changing the direction of the tension in the rope and therefore altering the net force acting on the system. As the angle increases, the acceleration may decrease due to a decrease in the net force.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
704
  • Introductory Physics Homework Help
3
Replies
102
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
941
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top