block A has a mass of 20kg and block B has a mass of 10kg. What is the value of static friction would stop the block from sliding?
5 answers
The value of static friction depends on the coefficient of static friction between the two surfaces in contact. Without knowing this coefficient, we cannot calculate the value of static friction.
block A has a mass of 10kg and block B has a mass of 25kg. Assume the coefficient of friction between the blocks is four times the coefficient of friction between the block and the plane. Determine the normal force between block B and the plane
Assuming that the surface between block B and the plane is horizontal and there is no acceleration:
The force of friction between block B and the plane can be calculated as:
Frictional force = coefficient of friction x normal force
Let n be the normal force between block B and the plane.
The weight of block B is:
Weight = mass x acceleration due to gravity
Weight = 25 kg x 9.81 m/s^2
Weight = 245.25 N
Since the surface is horizontal, the normal force is equal in magnitude to the weight of block B. Therefore, n = 245.25 N.
The force of friction between block B and the plane is:
Frictional force = 4 x coefficient of friction x n
Frictional force = 4 x coefficient of friction x 245.25 N
Since there is no acceleration, the force of friction must be equal and opposite to the force applied on block B by block A. This force is equal to the weight of block A:
Force on block B = Force on block A = weight of block A
Force on block A = mass of block A x acceleration due to gravity
Force on block A = 10 kg x 9.81 m/s^2
Force on block A = 98.1 N
Therefore,
Frictional force = 98.1 N
4 x coefficient of friction x 245.25 N = 98.1 N
coefficient of friction = 0.1
Thus, the value of the coefficient of friction between block B and the plane is 0.1.
The force of friction between block B and the plane can be calculated as:
Frictional force = coefficient of friction x normal force
Let n be the normal force between block B and the plane.
The weight of block B is:
Weight = mass x acceleration due to gravity
Weight = 25 kg x 9.81 m/s^2
Weight = 245.25 N
Since the surface is horizontal, the normal force is equal in magnitude to the weight of block B. Therefore, n = 245.25 N.
The force of friction between block B and the plane is:
Frictional force = 4 x coefficient of friction x n
Frictional force = 4 x coefficient of friction x 245.25 N
Since there is no acceleration, the force of friction must be equal and opposite to the force applied on block B by block A. This force is equal to the weight of block A:
Force on block B = Force on block A = weight of block A
Force on block A = mass of block A x acceleration due to gravity
Force on block A = 10 kg x 9.81 m/s^2
Force on block A = 98.1 N
Therefore,
Frictional force = 98.1 N
4 x coefficient of friction x 245.25 N = 98.1 N
coefficient of friction = 0.1
Thus, the value of the coefficient of friction between block B and the plane is 0.1.
block A has a mass of 10kg and block B has a mass of 25kg. Assume the coefficient of friction between the blocks is four times the coefficient of friction between the block and the plane. Determine the minimum coefficient of friction to prevent the blocks from slipping
To prevent the blocks from slipping, the force of friction between the two blocks must be greater than or equal to the force applied on block B by block A. This force is equal to the weight of block A:
Force on block B = Force on block A = weight of block A
Force on block A = mass of block A x acceleration due to gravity
Force on block A = 10 kg x 9.81 m/s^2
Force on block A = 98.1 N
The force of friction between the two blocks can be calculated as:
Frictional force = coefficient of friction between the blocks x normal force
Let n be the normal force between the two blocks. The normal force is equal in magnitude to the weight of block B:
n = weight of block B
n = mass of block B x acceleration due to gravity
n = 25 kg x 9.81 m/s^2
n = 245.25 N
Therefore, the force of friction between the blocks is:
Frictional force = 4 x (coefficient of friction between block B and the plane) x n
Frictional force = 4 x (1/4 x coefficient of friction between the blocks) x 245.25 N
Frictional force = coefficient of friction between the blocks x 245.25 N
To prevent the blocks from slipping, the frictional force must be greater than or equal to 98.1 N:
Frictional force ≥ 98.1 N
Substituting the equation for frictional force derived above:
coefficient of friction between the blocks x 245.25 N ≥ 98.1 N
coefficient of friction between the blocks ≥ 0.4
Therefore, the minimum coefficient of friction to prevent the blocks from slipping is 0.4.
Force on block B = Force on block A = weight of block A
Force on block A = mass of block A x acceleration due to gravity
Force on block A = 10 kg x 9.81 m/s^2
Force on block A = 98.1 N
The force of friction between the two blocks can be calculated as:
Frictional force = coefficient of friction between the blocks x normal force
Let n be the normal force between the two blocks. The normal force is equal in magnitude to the weight of block B:
n = weight of block B
n = mass of block B x acceleration due to gravity
n = 25 kg x 9.81 m/s^2
n = 245.25 N
Therefore, the force of friction between the blocks is:
Frictional force = 4 x (coefficient of friction between block B and the plane) x n
Frictional force = 4 x (1/4 x coefficient of friction between the blocks) x 245.25 N
Frictional force = coefficient of friction between the blocks x 245.25 N
To prevent the blocks from slipping, the frictional force must be greater than or equal to 98.1 N:
Frictional force ≥ 98.1 N
Substituting the equation for frictional force derived above:
coefficient of friction between the blocks x 245.25 N ≥ 98.1 N
coefficient of friction between the blocks ≥ 0.4
Therefore, the minimum coefficient of friction to prevent the blocks from slipping is 0.4.
4 answers