What is the minimum mass m that will stick and not slip?

In summary, to find the minimum mass m that will stick and not slip on the 20 degree slope, we need to use the coefficient of static friction (mu_s) and the weight of the hanging block (m2) to calculate the normal force and the maximum frictional force acting on m1. We can then set up equations using these forces and the tensions in the strings to solve for m1. The block is not accelerating, so all the force component equations will equal zero and the tensions in each diagram are equal. Additionally, the coefficient of kinetic friction is not needed for this calculation.
  • #1
mugzieee
77
0
a block of mass m is resting on a 20degree slope. The block has coefficients of friction mu_s =0.80 and mu_k =0.50 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg. (sorry didnt know of a way to distribute a picture)
what is the minimum mass m that will stick and not slip?

i drew 1 free body diagrams of the mass m and mass 2.
For the FBD of m1:
I have normal force, kinetic friction,static friction, the weight components of the mass, and the force of the second mass acting on the first one.

For the FBD of m2:
I have just the force acting on ms from m1, and the weight of m2.

im not quite sure if the block is accelerating kuz it says resting, but then again I am given coefficients of frictional forces. also can i equate the 2 tension forces of the ropes from the third law?
im pretty stumped, I am not sure how to proceed after drawing the FBD's.
 
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  • #2
The block is not accelerating, the coefficient of kinetic friction is not required and is likely only given to see if you understand that it is not needed. The coefficient of static friction will be required to calculate the minimum mass.

Because there will be no acceleration for the mass you are calculating, all the force component equations from your FBD's will equal zero, and the tensions in each diagram are equal.

Remember that the maxium frictional force (while the block is still at rest) is mu_s*N.
 
  • #3


To determine the minimum mass m that will stick and not slip, we need to consider the forces acting on the block of mass m. The block is resting on a 20 degree slope and is connected to a hanging block of mass 2.0 kg via a massless string over a frictionless pulley. The two blocks are also in contact with a surface that has coefficients of friction mu_s = 0.80 and mu_k = 0.50.

First, we need to determine if the block is in a state of equilibrium or if it is accelerating. Since the block is resting on the slope, we can assume that it is in a state of equilibrium. This means that the sum of all the forces acting on the block must be equal to zero.

From the free body diagram of the block, we can see that there are four forces acting on it: normal force, kinetic friction, static friction, and the force of the second mass acting on the first one. The normal force is perpendicular to the surface and is equal to the weight component of the block in the y-direction. The kinetic friction is acting in the opposite direction of the block's motion, which is down the slope. The static friction is acting in the opposite direction of the impending motion, which is up the slope. The force of the second mass acting on the first one is equal to the tension in the string, which is also pulling the block up the slope.

We can set up the equations of equilibrium to determine the minimum mass m that will stick and not slip. Since the block is in equilibrium, the sum of the forces in the x-direction and the sum of the forces in the y-direction must be equal to zero.

In the x-direction, we have:
T - mu_k*N = 0

In the y-direction, we have:
N - mg*cos(20) = 0

Using the given coefficients of friction and the weight of the block, we can solve for the normal force and tension in the string:
N = mg*cos(20)
T = mu_k*N = 0.50*mg*cos(20)

Now, we can substitute these values into the equation for the sum of the forces in the x-direction to solve for the minimum mass m:
T - mu_k*N = 0
0.50*mg*cos(20) - (0.50*mg*cos(20))*cos(20) = 0
m = (0
 

Related to What is the minimum mass m that will stick and not slip?

What is the minimum mass m that will stick and not slip?

The minimum mass m that will stick and not slip is determined by the coefficient of friction between the two surfaces in contact. This coefficient of friction is a measure of the resistance to sliding between two objects and is influenced by factors such as surface roughness and material properties.

How is the coefficient of friction determined?

The coefficient of friction is determined experimentally by measuring the force required to start an object in motion on a surface, and dividing it by the force pressing the two surfaces together. This ratio is known as the coefficient of friction.

What happens if the coefficient of friction is too low?

If the coefficient of friction is too low, the minimum mass m needed to prevent slipping will also be low. This means that even a small force or disturbance could cause the object to slip, making it unstable and potentially dangerous.

What is the role of surface roughness in determining the minimum mass m?

The roughness of the surfaces in contact plays a significant role in determining the minimum mass m that will stick and not slip. Rougher surfaces have a higher coefficient of friction, meaning that a larger mass will be needed to prevent slipping.

Can the minimum mass m change over time?

Yes, the minimum mass m can change over time due to factors such as wear and tear on the surfaces, changes in temperature, or the introduction of lubricants. It is important to regularly reassess the minimum mass m to ensure safety and stability of objects in contact.

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