High School Physics : Understanding Frictional Force

Study concepts, example questions & explanations for High School Physics

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Example Questions

Example Question #3 : Specific Forces

Susan is trying to push a \displaystyle 12kg crate across the floor. She observes that the force of friction between the crate and the floor is \displaystyle -50N. What is the coefficient of static friction?

\displaystyle \small g=-9.8\frac{m}{s^2}

Possible Answers:

\displaystyle 0.23

\displaystyle 0.56

It cannot be determined

\displaystyle 0.43

\displaystyle 0.11

Correct answer:

\displaystyle 0.43

Explanation:

The equation for the force of friction is \displaystyle F_{friction}=\mu F_{normal}, where \displaystyle \mu is the coefficient of static friction.

The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative of the force of gravity).

\displaystyle F_{normal}=-F_{gravity}=-(mg)

\displaystyle F_{normal}=-((12kg)(-9.8\frac{m}{s^2}))

\displaystyle F_{normal}=-(-117.6N)

\displaystyle F_{normal}=117.6N

Since the problem tells us that the force due to friction is \displaystyle -50N, we can plug these values into our original equation to solve for the coefficient of friction.

\displaystyle F_{friction}=\mu F_{normal}

\displaystyle (-50N)=\mu (117.6N)

\displaystyle \frac{-50N}{117.6N}=\mu

\displaystyle 0.43=\mu

The coefficient of friction has no units.

Example Question #1 : Understanding Frictional Force

Mary is trying to push a \displaystyle 12kg crate across the floor. She observes that the coefficient of static friction between the floor and the crate is equal to \displaystyle 0.12. What will be the force of friction on the crate?

\displaystyle \small g=-9.8\frac{m}{s^2}

Possible Answers:

\displaystyle -14.11N

\displaystyle -9.81N

\displaystyle 0N

\displaystyle -0.32N

\displaystyle -11.23N

Correct answer:

\displaystyle -14.11N

Explanation:

The equation for the force of friction is \displaystyle F_{friction}=\mu F_{normal}, where \displaystyle \mu is the coefficient of static friction.

The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative the force of gravity).

\displaystyle F_{normal}=-F_{gravity}=-(mg)

\displaystyle F_{normal}=-((12kg)(-9.8\frac{m}{s^2}))

\displaystyle F_{normal}=-(-117.6N)

\displaystyle F_{normal}=117.6N

Since the problem tells us that the coefficient due to friction is equal to \displaystyle 0.12, we can plug these values into our original equation to solve for the force.

\displaystyle F_{friction}=\mu F_{normal}

\displaystyle F_{friction}=(0.12)(117.6N)

\displaystyle F_{friction}=14.11N

As the force due to friction is moving in the OPPOSITE direction to the force Mary will exert, it should appropriately be labelled as negative.

 \displaystyle F_{friction}=-14.11N

Example Question #1 : Understanding Frictional Force

\displaystyle 2.9kg crate slides across a floor for \displaystyle 5s before coming to rest \displaystyle 10m from its original position.

What is the coefficient of kinetic friction on the crate?

\displaystyle \small g=-9.8\frac{m}{s^2}

Possible Answers:

\displaystyle 0.098

\displaystyle 0.8

\displaystyle 0.082

\displaystyle 1.6

\displaystyle 0.004

Correct answer:

\displaystyle 0.082

Explanation:

The equation for the force due to friction is \displaystyle F_f=\mu F_N, where \displaystyle \mu is the coefficient of kinetic friction. Since there is only one force acting upon the object, the force due to friction, we can find its value using the equation \displaystyle F=ma. We can equate these two force equations, meaning that \displaystyle ma=\mu F_N. We can solve for the normal force, but we need to find \displaystyle ma in order to find \displaystyle \mu.

The problem gives us the mass of the crate, but we have to solve for the acceleration.

Start by finding the initial velocity. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

\displaystyle \Delta x=\frac{(v_f+v_i)}{2}\Delta t

Plug in our given values and solve.

\displaystyle 10m=\frac{(0\frac{m}{s}+v_i)}{2}5s

\displaystyle \frac{10m}{5s}=\frac{v_i}{2}

\displaystyle 2\frac{m}{s}=\frac{v_i}{2}

\displaystyle 4\frac{m}{s}=v_i

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

\displaystyle \Delta x=v_it+\frac{1}{2}at^2

Plug in the given values to solve for acceleration.

\displaystyle 10m=(4\frac{m}{s}) (5s)+\frac{1}{2}a(5s)^2

\displaystyle 10m=(20m)+\frac{1}{2}a(25s^2)

\displaystyle -10m=\frac{1}{2}a(25s^2)

\displaystyle -20m=a(25s^2)

\displaystyle \frac{-20m}{25s^2}=a

\displaystyle -0.8\frac{m}{s^2}=a

Now that we have the acceleration and the mass, we can return to our first equation for force.

\displaystyle F=ma=\mu F_N

The normal force is the same as the mass times gravity.

\displaystyle ma=\mu (mg)

In this format, the masses cancel on both sides of the equation/

\displaystyle a=\mu * (g)

Now we can plug in our value for acceleration and gravity to solve for the coefficient of friction.

\displaystyle -0.8\frac{m}{s^2}=\mu(-9.8\frac{m}{s^2})

\displaystyle \frac{-0.8\frac{m}{s^2}}{-9.8\frac{m}{s^2}}=\mu

\displaystyle 0.082=\mu

 

Example Question #2 : Understanding Frictional Force

Erin pushes a \displaystyle 38kg cabinet across the floor. If it requires \displaystyle 171N of force, what is the coefficient of kinetic friction?

\displaystyle g=-9.8\frac{m}{s^2}

Possible Answers:

\displaystyle 0.214

\displaystyle 4.68

\displaystyle 0.372

\displaystyle 2.18

\displaystyle 0.46

Correct answer:

\displaystyle 0.46

Explanation:

The problem gives us the minimum force required to move the cabinet. That means the force Erin exerts will be equal to the force due to friction, but moving in the opposite direction. 

\displaystyle F=-F_f.

From here, expand the right side using the formula for kinetic friction and normal force.

\displaystyle F=-(\mu*F_N)

\displaystyle F=-\mu(mg)

Use the given values for the force, mass, and acceleration of gravity to solve for the coefficient of friction.

\displaystyle 171N=-\mu(38kg*-9.8\frac{m}{s^2})

\displaystyle 171N=-\mu(-372.4N)

\displaystyle \frac{171N}{-372.4N}=-\mu

\displaystyle -0.46=-\mu

\displaystyle 0.46=\mu

Example Question #1 : Understanding Frictional Force

It takes \displaystyle 20N of force to move a \displaystyle 19kg block from rest. Assuming no outside forces act upon the block, what is the coefficient of static friction?

Possible Answers:

\displaystyle 0.21

\displaystyle 9.31

\displaystyle 0.10

\displaystyle 0.98

\displaystyle 0.11

Correct answer:

\displaystyle 0.11

Explanation:

The force due to friction on a level surface is the product of the normal force and the coefficient of friction:

\displaystyle F_f=\mu_{s}F_N

We are told the force required to move the block, which will be equal to the force of friction, and the mass of the block. Use the mass of the block to calculate the normal force.

\displaystyle F_N=mg

\displaystyle F_N=(19kg)(9.8\frac{m}{s^2})=186.2N

Use the normal force and the force of friction to solve for the coefficient of friction.

\displaystyle F_f=\mu_{s}F_N

\displaystyle 20N=\mu_s(186.2N)

\displaystyle \mu_s=\frac{20N}{186.2N}

\displaystyle \mu_s=0.11

 

Example Question #1 : Understanding Frictional Force

Which of the following cannot be true of an object on a given surface?

Possible Answers:

Static friction: \displaystyle 10N

Kinetic friction: \displaystyle 3N

Static friction: \displaystyle 5N

Kinetic friction: \displaystyle 3N

Static friction: \displaystyle 8N

Kinetic friction: \displaystyle 5N

Static friction: \displaystyle 10N

Kinetic friction: \displaystyle 8N

Static friction: \displaystyle 5N

Kinetic friction: \displaystyle 8N

Correct answer:

Static friction: \displaystyle 5N

Kinetic friction: \displaystyle 8N

Explanation:

Kinetic friction is never greater than static friction. More force is always requires to overcome static friction than is required to overcome kinetic friction. It can require a large force to initiate motion, causing an initial acceleration by overcoming static friction. Once motion has begun, however, less for is required to maintain the motion due to the principles of Newton's first law and inertia.

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