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Calculating Friction

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# Calculating the values of friction forces acting between the surfaces of two bodies in contact with each other, using friction formula and equation, and applying the concept of static friction.

Question 1:

A body of mass 5kg is placed on a horizontal plane. It is found that a force of 10N applied horizontally to the body is just about to move it. Calculate the coefficient of static friction between the body and the plane.

Solution

Given: since the force of 10N was applied when the body was just about to move, then the static or limiting frictional force, F = 10N

The normal reaction R on the body is equal to its weight.

i.e., weight = mass x acceleration due to gravity (10m/s2)

R = W

= 5 x 10

= 50 Newtons

Therefore, from the formula of friction,

F = µ R

µ = F/R

= 10/50

= 0.2

The coefficient of static friction between the body and the plane is 0.2

Question 2:

A body of mass 40kg is given an acceleration of 10m/s2 on horizontal ground for which the coefficient of friction is 0.5. Calculate the force required to accelerate the body. (Take g = 10m/s2)

Solution

Notice that the body is accelerated, which means the force that acted on it is more than the limiting frictional force.

Therefore, the equation of motion of the body is given as:

F - µR = ma

F represents the applied force

µR represents the force of friction

ma represents the net or resultant force

F = ma + µR

= 40 x 10 + (0.5 x 40 x 10)

= 400 + 200

= 600N

The force required to accelerate the body is 600 Newtons.

Question 3:

A body of weight W is placed on a rough inclined plane with angle of inclination30o. Calculate the coefficient of static friction if the body is just on the point of slipping.

Solution

The diagram above shows a body on an inclined plane and the different forces acting on it.

R represents the normal reaction

F represents the static or limiting frictional force

W represents the weight of the body.

Now, we have to resolve the forces along the plane for equilibrium, and perpendicular to the plane.

Resolving along the plane for equilibrium:

F = Wsinθ  --- equation (1)

Resolving perpendicular to the plane:

R = Wcosθ  --- equation (2)

Since at the point of sliding, frictional force is proportional to the normal reaction, giving the equation

F = µR

Therefore, µ = F/R

Now dividing equation (1) above by equation (2), we have

F/R = Wsinθ/Wcosθ = µ

µ = Wsinθ/Wcosθ

µ = tanθ

Therefore, the coefficient of friction, µ for a body on an inclined plane is always given as the tangent of the angle of inclination.

Answering the question: the coefficient of friction of the body on an inclined plane with angle of inclination 30o is

Tan30o = 1/3  = 0.58

Question 4:

A body of mass 4kg is on the point of slipping down a plane, which is inclined at 30o to the horizontal. What force, parallel to the plane, will just move it up the plane? (Take g = 10m/s2).

Solution

Let P be the force needed to move the body up the plane, and F = µR the limiting frictional force opposing the motion. Notice that the difference of the forces on the body along the plane is P - µR since P will be greater than the frictional force µR, for the body to be moved up.

Resolving along the plane for equilibrium:

Sin30o  =  (P – µR)/W

P – µR = Wsin30o

P = µR + Wsin30o --- equation (1)

Resolving perpendicularly to plane:

Cos30o = R/W

R = Wcos30o  --- equation (2)

Substituting Wcos30o for R in equation (1), we have

P = µWcos30o + Wsin30o

Remember that              µ = tanθ

Therefore,                  µ = tan30o = 1/3

P = 1/3 x W x 3/2 + W x 1/2

P = W/2 + W/2

P = W

W = weight of the body = m x g

= 4 x 10

= 40N

Therefore, the force that is needed to just move the body up the plane is 40 Newtons.