Download Coefficient of Friction Practice

Friction Practice
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Complete on a separate sheet of paper, do not cram your answers in these spaces!
Friction Problems with Constant Speed Motion
1) The largest flowers in the world are the Rafflesia arnoldii, found in Malaysia. A single flower is
almost a meter across and has a mass up to 11.0 kg. Suppose you cut off a single flower and drag
it along the flat ground. If the coefficient of kinetic friction between the flower and the ground is
0.39, what is the magnitude of the frictional force that must be overcome?
FN
Pulling force must be at least equal to kinetic friction force (should be even bigger
to start actual motion).
Ff=µFN
(definition of friction force)
FN–Fg =0 or FN=Fg
(no acceleration vertically)
Ff
Fg=mg
(definition of gravity force)
So, substituting gives Ff=µmg=
= 0.39 · 11.0 kg · 9.8 N/kg = 42 N
Fg
2) Robert Galstyan, from Armenia, pulled two coupled railway wagons a distance of 7 m using his
teeth. The total mass of the wagons was about 2.20 × 105 kg. Of course, his job was made easier
by the fact that the wheels were free to roll. Suppose the wheels are blocked and the coefficient of
static friction between the rails and the sliding wheels is 0.220. What would be the magnitude of
the minimum force needed to move the wagons? Assume that the track is horizontal.
Note: 7 m is not needed to answer the question (it is just what he did).
Minimum force=Ff=µmg= 0.220 · 2.20 · 105 kg · 9.8 N/kg = 4.7 · 105 N
3)
A 24 kg crate initially at rest on a horizontal floor requires 75 N horizontal force to set it in
motion. Find the coefficient of static friction between the crate and the floor.
Ffs=µsmg
(subscript 's' refers to 'static')
µs=Ffs/mg= 75 N / (24 kg · 9.8 N/kg) = 0.32
4) Once the crate in the above problem is in motion, a horizontal force of 53 N keeps the crate
moving with a constant velocity. Find µk, the coefficient of kinetic friction, between the crate and
the floor.
No acceleration, so F=Ffk
(subscript 'k' refers to 'kinetic')
Ffk=µkmg
µk=Ffk/mg= 53 N / (24 kg · 9.8 N/kg) = 0.23
5)
A 25 kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in
motion. Once the chair is in motion, a 127 N horizontal force keeps it moving at a constant
velocity.
a) Find the coefficient of static friction between the chair and the floor.
As in #3:
µs=Ffs/mg= 165 N / (25 kg · 9.8 N/kg) = 0.67
b) Find the coefficient of kinetic friction between the chair and the floor.
As in #4:
µk=Ffk/mg= 127 N / (25 kg · 9.8 N/kg) = 0.52
6)
The coefficients of static and kinetic friction between wood and concrete are 0.60 and 0.45.
a) How much force would be needed to start a 45 kg wooden crate moving across a concrete
floor?
Fneeded=Ffs=µsmg=0.60 · 45 kg · 9.8 N/kg = 264.6 N ≈ 270 N
(2 significant figures)
b) How much force would be needed to keep it moving at constant velocity?
Fneeded=Ffk=µkmg=0.45 · 45 kg · 9.8 N/kg = 198.45 N ≈ 2.0 · 102 N (2 significant figures)
Friction Problems with Accelerated Motion
7) A force of 40.0 N accelerates a 5.0 kg block across the floor at 6.0 m/s 2. How big is the frictional
force on the block? What is the coefficient of friction between the block and floor?
Did in class.
F – Ff = ma
(h)
FN – F g = 0
(v)
Also, Ff = µFN
(f)
Also, Fg = mg
(g)
From (h): a = (F – Ff)/m
Substituting from (f): a = (F – µFN)/m
From (v) and (g): FN=Fg=mg
So, substituting into expression for a:
a = (F – µmg)/m = (710 N – 0.20 · 225 kg · 9.8 N/kg) / 225 kg =
= 1.2 N/kg = 1.2 m/s2
FN
A 225 kg crate is pushed horizontally across the floor with a force of 710 N. If the coefficient of
kinetic friction is 0.20, find the acceleration of the crate.
Ff
F
Fg
8)
9) A 5.0 kg block of cheese is accelerated horizontally across a table top at a rate of 1.2 m/s 2 as it is
pushed onto a conveyor system in a cheese factory. If the coefficient of friction between the
cheese and table is 0.33, what amount of force is necessary to do this?
F – Ff = ma
Note: 1.2 m/s2 = 1.2 N/kg
F = Ff + ma = µmg + ma = m(µg + a) = 5.0 kg (0.33 · 10 N/kg + 1.2 N/kg) =
= 5.0 kg (3.3 N/kg + 1.2 N/kg) = 5.0 kg · 4.5 N/kg = 22.5 N
Note: this time, the value for 'g' was chosen to be 10 N/kg instead of 9.8 N/kg
to simplify calculations (Dr. K did it in his head)
10) A 10.0 N force is applied to the side of a 4.0 kg block that is sitting on a table. If the coefficient of
kinetic friction is 0.20, determine the acceleration of the block.
Similar to #8.
a = (F – µmg)/m = (10.0 N – 0.20 · 4.0 kg · 10 N/kg) / 4.0 kg = 0.5 N/kg = 0.5 m/s2
Friction Problems with Applied Force Acting at Angles
11) A 22 kg sled is pulled across the ground at a constant speed with 15 N of force by a rope making a
35° angle above the horizontal. What is the coefficient of friction between the sled and the
ground?
12) A 22 kg sled is pulled across the ground with 45 N of force by a rope making a 35° angle above
the horizontal. If the coefficient of friction between the sled and the ground is 0.15, what is the
acceleration of the sled?
13) Manny is pushing a 5.0 kg box across a level floor with a force of 50.0 N. The force he applies is
downward and to the right at 25°. The coefficient of kinetic friction between the box and the floor
is 0.22. Find the acceleration of the box.
14) A box of books weighing 325 N moves at a constant velocity across the floor when the box is
pushed with a force of 425 N exerted downward at an angle of 35.2° below the horizontal. Find µk
between the box and the floor.
Additional Practice - Mixed types
15) A blue whale with a mass of 1.90 × 105 kg was caught in 1947.What is the magnitude of the
minimum force needed to move the whale along a horizontal plank if the coefficient of static
friction between the plank’s surface and the whale is 0.460?
16) A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s 2. How big is the
frictional force on the block? What is the coefficient of friction between the chair and floor?
17) Until 1979, the world’s easiest driving test was administered in Egypt. To pass the test, one
needed only to drive about 6 m forward, stop, and drive the same distance in reverse. Suppose
that at the end of the 6 m the car’s brakes are suddenly applied and the car slides to a stop. If the
frictional force required to stop the car is 6.0 × 103 N and the coefficient of kinetic friction
between the tires and pavement is 0.77, what is the magnitude of the car’s normal force? What is
the car’s mass?
18) A 4.00 kg block is pushed along the floor with a constant applied force of 85.0 N that acts at an
angle of 55.0 degrees below the horizontal. The block accelerates to the right at 6.00 m/s 2.
Determine the coefficient of kinetic friction between the block and then floor.
19) A 425 kg steel crate is pushed horizontally across a steel beam with a force of 3600 N. If the
coefficient of kinetic friction of steel on steel is 0.57, find the acceleration of the crate.
20) A student pulls on a rope attached to a box of books and moves the box down the hall. The
student pulls with a force of 185 N at an angle of 25.0° above the horizontal. The box has a mass
of 35.0 kg, and µk between the box and the floor is 0.27. Find the acceleration of the box.