Download Friction Problems ACTIVITY 1: Cut out the problem and the steps

Friction Problems
ACTIVITY 1: Cut out the problem and the steps. Rearrange the steps and paste on a
piece of paper, with the question, so that the solution is in the right order.
EXAMPLE 1: A 30.0g arrow is shot by William Tell through an 8.00cm thick apple
sitting on top of his son’s head. If the arrow enters at 30.0m/s and emerges at 25.0m/s,
in the same direction, with what force has the apple resisted the arrow?
In order to determine force, we need to multiply mass by acceleration. However, we do
not know acceleration, so we must calculate it using the time independent velocity
formula from the linear motion unit:
2
2
v f  v o  2a(d)
25.0 2  30.0 2  2a(0.0800)
625  900  0.16a
900  900
.
275 0.16a

0.16
0.16
a  1718.75m /s2  1720m /s2
Thus we find that the apple applied a force of -51.6N on the arrow as it passed through
the apple.
 the resistance force on a 30.0g arrow, exerted by an 8.00cm
We want to determine
apple, that caused the arrow to decelerate from 30.0m/s to 25.0m/s.
Now that we know the acceleration and the mass, we can calculate the force using
F=ma:
F  ma
.
F  (0.0300)(1720)  51.6N
To begin, we first need to make sure all our units are standard. Converting grams to
kilograms, we get that the mass of the arrow is 0.0300 kg. Converting centimeters to
meters, we get thatthe width of the apple is 0.0800m.
EXAMPLE 2: Rose is sledding down an ice-covered hill inclined at an angle of 15O to the
horizontal. If Rose and the sled have a combined mass of 54.0kg, what is the force
pulling them down the hill?
The weight of the sledder and sled, combined, can be calculated using the formula
Fg=mg:
Fg  mg  (54.0)(9.8)  529.2N
Thus, we find that the force pulling Rose and her sled down the hill is 137 N.
We want to determine the component of the force of gravity acting on Rose and the
sled in the directionof the incline of the hill.
We must now find the component of this weight in the direction of the slope. To
visualize the vector components in this problem, we must draw a diagram and define a
reference frame on that diagram. Notice that the “horizontal” direction of our
reference frame is parallel to the slope of the hill. In this reference frame, the direction
of the weight vector is actually 90O-15O =75O, or the compliment of the angle of
inclination:
We know the angle of incline of the hill is 15O and the mass of Rose and the sled
combined is 54.0kg.
We see that the component we want is the horizontal component, so we use the
formula for the horizontal component of a vector:
Fx  F cos  529.2(cos 75O )  136.97N  137N

Example 3: A 300kg crate is at rest on a ramp that is inclined at 30O to the horizontal.
What is the coefficient of static friction between the crate and the ramp?
In this equation, we know that the mass is 300kg, that g=9.8m/s2 and that the angle is
the compliment of the ramps angle of inclination, so and that the angle is the
compliment of the ramps angle of inclination, so θ=60O. But, we don’t know what the
normal force is. To calculate this, we need to go back to the free-body diagram, which
tells us:
.
We want to determine the coefficient of static friction between a 300kg crate that is at
rest on a ramp that is inclined at 30O to the horizontal.
Now, we can go back and solve the previous equation for the coefficient of friction:
.
By substitution, we get:
.
Thus, the coefficient of static friction between the crate and the ramp is 0.58.
We see from the free-body diagram that the force of friction is equal to the x-component
of the crate’s weight, and with substitution we can get a formula that contains the
coefficient of friction:
This is essentially a static equilibrium problem, so we must begin by drawing a freebody diagram and creating the up=down and left=right equations:
Example 4: At Sea World, a 900 kg Polar Bear slides down a wet slide inclined at an
angle of 25.0O to the horizontal. The coefficient of friction between the polar bear and
the slide is 0.0500. What frictional force impeded the bears motion down the slide?
Calculating the bear’s weight we get:
Fg  mg  900(9.8)  8820N .
We want to calculate the force of friction acting on a 900 kg bear that is sliding down a
25.0O incline with a coefficient of friction of 0.0500.
 force of friction we get:
Finally, calculating the
Ff  FN  0.0500(7994)  400N .
Calculating the vertical component of the weight to determine the normal force we get:
FN  Fgy  Fg sin   8820(sin 65)  7994N .
 the force of friction, we must first determine the normal force
In order to determine
acting on the bear by the incline surface.
 of friction acting on the bear is 400N.
Thus, the force
In the diagram below, you can see that the normal force equal to the vertical
component of the bear’s weight. However, the direction of the weight vector is the
compliment of the angle of incline, so the direction is 65O:
Activity 2: Solve each problem and write a formal solution for each.
1. Carter’s favorite ride at Playland Amusement Park is the rollercoaster. The
rollercoaster car and passengers have a combined mass of 1620 kg and descend
the first hill at an angle of 45.0O to the horizontal. With what force is the
rollercoaster pulled down the hill?
2. Unbeknownst to the students, every time the school floors are waxed, Mr.
Swafford likes to slide down the hallway in his socks. Mr. Swafford weighs
850N and the coefficient of sliding friction between his socks and the floor is
0.120. What is the force of friction that opposes Mr. Swafford’s motion down the
hall?
3. Skye is trying to make her 70.0kg Saint Bernard go out the back door, but the dog
refuses to walk. If the coefficient of sliding friction between the dog and the floor
is 0.50, how hard must Skye push in order to move the dog with a constant
speed?
4. A child’s wagon of mass 20kg is rolling across the grass with initial speed of
10m/s. The force of friction slows it to a stop in 5s. What is the force of rolling
friction between the wheels of the wagon and the grass?
5. Rather than taking the stairs, Martin gets from the second floor of his house to
the first floor by sliding down the banister, which is inclined at an angle of 30.0O
to the horizontal. A) If Martin has a mass of 45kg and the coefficient of sliding
friction between Martin and the banister is 0.20, what is the force of friction
impeding Martin’s motion down the banister? B) If the banister is made steeper,
what effect, if any, would this have on the force of friction?
6. A 1250kg hippo slides down a mud-covered hill inclined at an angle of 18O to the
horizontal. A) If the coefficient of sliding friction between the hippo and the
mud is 0.0900, what force of friction impedes the hippo’s motion down the hill?
B) If the hill were steeper, how would this affect the coefficient of sliding friction?
C) If the hill were steeper, how would this affect the force of sliding friction?
7. Noah is loading the ark, but one elephant refuses to budge up the ramp, causing
a traffic jam for the animals in line behind it. If the 1500kg elephant is at rest on
the ramp, which is inclined at 10O to the horizontal, then what is the coefficient of
static friction between the elephant and the ramp?