Friction with two masses (Help)

In summary, there is a slab of mass 40 kg and a block of mass 10 kg resting on top of it. The coefficient of static friction between the block and slab is 0.6 and the coefficient for kinetic friction is 0.4. The block is pulled to the right with a force of 100 N. The resulting accelerations of the block and the slab can be calculated using the equation F=ma, where F is the total force acting on the system and m is the mass of the respective object. The acceleration of the top block is 6.1 m/s^2, but the acceleration of the bottom slab cannot be determined without drawing free body diagrams for each mass and considering the equal and opposite
  • #1
DJWise
3
0
1. A slab of mass 40 kg rest on a frictionless floor, and a block of mass 10kg rest on top of the slab. Between block and slab, the coefficient of static friction is 0.6, and the coefficient for kinetic friction is 0.4. The block is pulled by a horizontal force (to the right) of magnitude 100N. In unit vector notation, what are the resulting accelerations of a.) the block and b.) the slab?



2. F=ma



3. Some how got the top mass to be accelerating 6.1 m/s^2 but can't get what the bottom slab is accelerating at
 
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  • #2
DJWise said:
1. A slab of mass 40 kg rest on a frictionless floor, and a block of mass 10kg rest on top of the slab. Between block and slab, the coefficient of static friction is 0.6, and the coefficient for kinetic friction is 0.4. The block is pulled by a horizontal force (to the right) of magnitude 100N. In unit vector notation, what are the resulting accelerations of a.) the block and b.) the slab?



2. F=ma



3. Some how got the top mass to be accelerating 6.1 m/s^2 but can't get what the bottom slab is accelerating at


I agree with your answer for the acceleration of the top block.

Have you drawn Free Body Diagrams for each mass? If not, do so (separate diagrams for each), making sure to show all forces acting on each one.

If you look at how you got that answer, one component should include a frictional force. Remember that a frictional force will have an equal and opposite force on each of the surfaces on which it acts.

Also remember that the force acting on each mass independently will be equal to the total force acting on the system (100 N to the right).
 
  • #3


I would first like to clarify the setup of the problem. The two masses, the slab and the block, are in contact with each other and are on a frictionless floor. This means that there is no external force acting on the system in the vertical direction. The only forces acting on the system are the horizontal force of 100N pulling the block and the frictional force between the two masses.

To solve this problem, we can use the equation F=ma, where F is the net force acting on the object, m is the mass of the object, and a is the resulting acceleration. Since the block and the slab are in contact with each other, they will have the same acceleration in the horizontal direction.

a.) To find the acceleration of the block, we can use the equation F=ma and plug in the values given in the problem. Since the block is being pulled to the right, the resulting acceleration will also be in the right direction. So, we have:

100N = (10kg)a

a = 10m/s^2

Therefore, the block will have an acceleration of 10m/s^2 in the right direction.

b.) To find the acceleration of the slab, we can use the same equation, but with a slight modification. Since the slab is in contact with the block, it will experience a frictional force in the opposite direction of the block's motion. So, we have:

F - μmg = ma

Where F is the applied force, μ is the coefficient of static friction, m is the mass of the block, and a is the acceleration of the slab. Plugging in the values given in the problem, we have:

100N - (0.6)(10kg)(9.8m/s^2) = (40kg)a

a = 6.1m/s^2

Therefore, the slab will have an acceleration of 6.1m/s^2 in the right direction.

In summary, the resulting accelerations of the block and the slab are 10m/s^2 and 6.1m/s^2, respectively, both in the right direction. It is important to note that the acceleration of the block is greater than that of the slab, as the block is being pulled by an external force while the slab is only experiencing a frictional force.
 

1. What is friction with two masses?

Friction with two masses refers to the resistance force that occurs when two objects with mass are in contact and moving relative to each other. It is caused by the microscopic irregularities on the surface of the objects rubbing against each other.

2. How does the mass of the objects affect friction?

The mass of the objects does not directly affect friction. However, the weight of the objects, which is affected by mass, can have an indirect effect on the amount of friction. Heavier objects have greater normal force, which can increase friction.

3. What factors affect the amount of friction between two masses?

The amount of friction between two masses is affected by factors such as the type of surfaces in contact, the force pressing the surfaces together (normal force), and the presence of any lubricants or contaminants between the surfaces.

4. How can friction between two masses be reduced?

Friction between two masses can be reduced by using lubricants, which reduce the contact between the surfaces and therefore decrease friction. Additionally, using smoother or more slippery surfaces can also reduce friction.

5. What is the difference between static and kinetic friction?

Static friction is the resistance force between two objects that are not moving relative to each other. It increases as the applied force increases, until the objects start moving. Kinetic friction, on the other hand, occurs when two objects are already in motion relative to each other. It is typically lower than static friction.

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