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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
UNIT 4: ANSWERS TO PROBLEMS
PROBLEM 1
Consider the following diagram showing the two
horizontal forces applied by two friends as they pull
a 1.20 × 103 kg car along an icy road.
a) Find the total force (in component form).
+y
F1 = 456 N
This is just another vector problem. We have to add
two vectors just like we would with displacement
vectors. The only difference is the units are N instead of m.
F2 = 642 N
2 =
50.0°
75.0°
+x
The variables are defined in the diagram that came with
the question. We need to find θ1 measured from the positive
x-axis in order to find the components of F1 .
+y
Using the geometry
θ
F1
θ1 = 180.0 º – 50.0º = 130.0°
F1y
50.0°
The components are then
+x
F1x
F1 y  F1 sin 1   456 N  sin130.0  349 N
F1x  F1 cos 1 =  456 N  cos130.0  293 N
+y
F2 y  F2 sin  2   642 N  sin 75.0  620 N
F2 x  F2 cos  2   642 N  cos 75.0  166 N
F2
F1y
75.0°
F1x
+x
The sum of the forces F12 is
Fnet  F1  F2 
 293 N  ˆi  349 N  ˆj  166 N  ˆi   620 N  ˆj
   293 N   166 N   ˆi    349 N    620 N   ˆj
  127 N  ˆi   969 N  ˆj
The net force is Fnet   127 N  ˆi   969 N  ˆj .
If the question had not specified the form of the force vector, we would have
continued to
calculate the angle and magnitude of the vector to get an answer in the same form as the
original question.
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
b)
Calculate the acceleration (magnitude and direction) of the car.
Besides the forces from part a), we have
m  1.20 103 kg, a  ?
Newton’s second law of motion states
Fnet  ma
Rearranging
a


Fnet
m
 127 N  ˆi   969 N  ˆj
1.20  103 kg
 127 N  ˆi
1.20  103 kg


 969 N  ˆj
1.20  103 kg
 

 0.1058 m/s 2 ˆi  0.8075 m/s 2 ˆj
Converting to magnitude – direction form
a
 0.1058 m/s    0.8075 m/s 
2
2
2
2
 0.814 m/s 2
 0.8075 m/s   180
  arctan
 0.1058 m/s 
2
2
 82.5  180
add 180° because x component is negative
 97.5
The car accelerates at 0.81 m/s2 at 97.5° from the x-axis
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
PROBLEM 2
Earth is attracted towards that 70 kg man with a force of 686 N. Why don’t we notice the effect of our
gravitational force on Earth?
Define Fg = 686 N, and the mass of the earth M = 5.981024 kg. If the noted force was the only force
acting on the earth, the acceleration of the earth would be
a

Fg
M
686 N
5.98  1024 kg
 1.1  1022 m/s 2
This is tiny. Starting at rest and accelerating at this rate for the entire age of the universe (~15 billion
years) would result in a final velocity of 510-5 m/s. When we include all of the other forces on the earth
(moon, sun, planets, galaxy, other people …), the force from one person is just a tiny part of the force on
the earth. The gravitational force from the Earth on us is large compared to most of the major forces
acting on us. The gravitational force from us on the Earth on us is small compared to most of the major
forces acting on the earth.
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
PROBLEM 3
Problem 3 and Page 12 Normal and Gravitational Forces on Figures
Draw the normal and gravitational forces in the following systems.
FN
FN
surface
FN
Fg
Fg
surface
surface
Fg
Problem 3
Find the acceleration of a 1600 kg car that is being pushed horizontally with a force of 5200 N to the
right, assuming that the coefficient of friction is 0.25.
Is there acceleration?
We expect acceleration to the right so we set the x-axis along that
direction, with the y-direction vertical.
y
FN
x
Fk
Draw FBD, define directions
v
Define terms: m  1600 kg,
F  5200 N ˆi,
k  0.25,
Fg
F
a ?
We have forces in all directions. The normal force is just enough to balance the force of gravity so there
is no acceleration in the y-direction. The equation for the forces in the x-direction is
Fnet , x  F  Fk  max
ax 
F  Fk
m
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
Here, the friction force will be against the direction of motion so the sign for the friction force is
included in the formula (the alternative is to add the friction force and give it a negative value, both
methods are valid).
In this expression, we know mass and applied force so we need to find a way to get the friction force.
We know that
Fk  k FN
and we can use the motion in the y-direction to find FN .
Fnet , x  FN  Fg  0
FN  Fg
 mg

 1600 kg  9.8 m/s 2

 1.568  104 N
Now, we can find the friction force.
Fk   k FN

  0.25  1.568  10 4 N

 3.92  103 N
The acceleration is then
ax 

F  Fk
m
 5200 N    3920 N 
1600 kg
 0.80 m/s 2
The acceleration is 0.80 m/s2 to the right (positive x-direction).
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
PROBLEM 4
A boat is secured to a lakeside pier with two horizontal ropes. A wind is blowing off shore. The tension
in the ropes are T1 = 48 N [16° N of E] and T2 = 48N [16° S of E]. Assuming that the net horizontal force
on the boat is zero, what is the force of the wind on the boat?
There is no movement or tension in the vertical direction so we will only look at the horizontal (E-N)
plane.
T1
Draw FBD, draw directions
Fw
16°
-16°
Terms defined in question.
Is the object accelerating?
N
no
What is the net force on the boat?
E
T2
0
The sums of the force components in both directions must be
.
0
Break the tension forces into components
T1E  T1 cos 1
T2 E  T2 cos  2
  48 N  cos16
  48 N  cos  16 
 46.1 N
 46.1 N
T1N  T1 sin 1
T2 N  T2 sin  2
  48 N  sin16
  48 N  sin  16 
 13.2 N
 13.2N
Use the net force to find the components of the wind force Fw .
Fnet  T1  T2  Fw  0
Rearranging
Fw  T1  T2
FwE  T1E  T2 E
FwN  T1N  T2 N
   46.14 N    46.14 N 
  13.2 N    13.2 N 
 93.3 N
0N
The force from the wind is –93 N to the east, or 93 N to the west.
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
PROBLEM 5
Problem 5
Imagine, instead of pushing the car in Problem 3, you are tow the same car (mass 1600 kg, coefficient of
friction 0.25) with a tow line that makes an angle of 37° with the horizontal. The tension on the tow line
is 5200 N. What is the acceleration of the car?
y
Draw the FBD
Define terms:
FN
x
T
m  1600 kg, T  5200 N ˆi 37 ,
k  0.25,
Fk
ax  ?
θ = 37°
v
Once again, we expect the acceleration to be in the x-direction only
(any acceleration in the y-direction would drive the car off the ground
or into the ground). The big difference from problem 3 is the tension
with components in both x and y-directions. All other forces are in single directions.
Fg
The net force in the x-direction is
Fnet , x  Tx  Fk  max
Rearranging,
ax 
T cos   Fk
,
m
We need to use the y-direction net force equation to find the normal force, and use that to find the
force of friction.
Fnet , x  FN  Ty  Fg  0
FN  Fg  Ty
 mg  T sin 


 1600 kg  9.8 m/s 2   5200 kg  cos 37 
 1.255  104 N
Now, we can find the friction force.
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
Fk   k FN

  0.25  1.255  10 4 N

 3.14  103 N
The acceleration is then
ax 

T cos   Fk
m
 5200 N  cos 37   3140 N 
1600 kg
 0.63 m/s 2
The acceleration is 0.63 m/s2 to the right (positive x-direction).
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
PROBLEM 6
Problem 6 – Hint
A child pulls a sled up a snow-covered hill at a constant velocity with a force parallel to the hillside. If
the sled has a mass of 9.5 kg, the hill slope is 12°, and the coefficient of friction between the sled and
the snow is 0.20, what is the force applied by the child?
There are several important notes to solving this problem.
Check out the hints.
y
FN
x
F
Draw FBD
Fk
Hint 1:
The normal force is always perpendicular to the
surface. In this case, it is not opposite the force
of gravity.
v
θ = 12°
Fgy
Fg
Fgx
Hint 2:
Choose directions as shown with the x in the direction of
motion (up the ramp) and y in the direction of the normal.
This greatly simplifies the math because we only have to break the force of gravity into
components. The other forces all lie entirely along the x or y in the direction.
We will need to find the components of the gravitational
force (see the diagram to the right to verify that the angle
between the y-direction and gravitational force vector is 12°).
y
x
θ = 12°
Fgy  mg cos 
Fgx  mg sin 
Hint 3:
Notice that the trig functions usually associated with the
x and y- components are switched. This comes from the
fact that the angle  is measured relative to the y-axis
instead of the x-axis.
θ = 12°
Fgy
Fg
Fgx
The set of hints here are steps common by all ramp problems. Ramp problems show up on assignments
and exams so become very familiar with the steps.
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
Problem 6 - Solution
A child pulls a sled up a snow-covered hill at a constant velocity with a force parallel to the hillside. If
the sled has a mass of 9.5 kg, the hill slope is 12°, and the coefficient of friction between the sled and
the snow is 0.20, what is the force applied by the child?
y
x
FN
Draw FBD
F
Hint 1:
The normal force is always perpendicular to the
surface. In this case, it is not opposite the force
of gravity.
Fk
v
θ = 12°
Fgy
Fg
Fgx
Hint 2:
Choose directions as shown with the x in the direction of
motion (up the ramp) and y in the direction of the normal.
This greatly simplifies the math because we only have to break the force of gravity into
components. The other forces all lie entirely along the x or y in the direction.
We will need to find the components of the gravitational
force (see the diagram to the right to verify that the angle
between the y-direction and the force of gravity vector is 12°).
y
x
θ = 12°
Fgy  mg cos 
Fgx  mg sin 
θ = 12°
Hint 3:
Notice that the trig functions usually associated with the
x and y- components are switched. This comes from the
fact that the angle  is measured relative to the y-axis
instead of the x-axis.
Fgy
Fg
Fgx
Is the object accelerating? No, the child is pulling the sled at a constant velocity, so net force is 0.
Define terms:
m  9.5 kg, k  0.20,   12,
Fnet  0,
F ?
We are interested in the force applied by the child so let’s look at the x-directions.
Fnet , x  F  Fgx  Fk  0
F  Fgx  Fk
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
We don’t know the friction force in this equation. This requires a look at the net force in the y-direction
so we can find the normal force, and then the force of friction.
Fnet , y  FN  Fgy  0
FN  Fgy
 mg cos 
The friction is then
Fk  k FN
 k mg cos 
Insert this into the x-component force equation
F  Fgx  Fk
 mg sin   k mg cos 




  9.5 kg  9.8 m/s 2 sin12   0.20  9.5 kg  9.8 m/s 2 sin12
 37.6 N
The force from the child is 38 N up the hill.
In this question, the actual calculations were left to the very end. This reduces the chance of having a
round off error. In other problem solutions, one or two extra significant figures are kept until the final
calculation. This is an old note but it doesn’t hurt to repeat it.
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
PROBLEM 7
Sleds A and B are connected by a horizontal rope, with A in front of B. Sled A is pulled forward with a
horizontal rope with a tension of magnitude 29.0 N. The masses of sleds A and B are 6.7 kg and 5.6 kg,
respectively. The magnitudes of kinetic friction on A and B are 9.0 N and 8.0 N respectively.
B
A
TAB
Tpull = 29.0 N i
mB = 5.6 kg, FkB  –8.0 N ˆi, mA  6.7 kg, FkA  –9.0 N ˆi
a) Find the acceleration of the two-sled system
y
x
Draw FBD or FBDs
choose directions
FNA,
FNB
FkA, FkB
Is the object accelerating?
yes – need to find a
Can we consider the sleds as one item?
Sure. The sleds move together, with the same acceleration.
There are cases where we would have to use two FBDs, like
if we didn’t have the friction forces, but a combined FBD
works in this case.
Tpull
v
FgA, FgB
The combined mass of the two sleds is
m  mA  mB
  5.6 kg    6.7 kg 
 12.3 kg
Do we need to worry about the net force in the y-direction?
In this case there is no movement in the y-direction (the net
force in the y-direction is
zero), and we don’t need to find the normal force to calculate the forces of friction. The
y-direction forces have no components in the x-direction so we can then ignore them.
The net force in the x-direction is the sum of the forces on the system. The directions of the friction
forces are already taken into account in the variable definitions so they are added (added negative
values instead of subtracted positive values.)
Fnet , x  Tpull  FkA  FkB  max
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
Solving for the acceleration in the x-direction (the total acceleration because there is none in the ydirection).
ax 

T pull  FkA  FkB
m
 29.0 N    9.0 N    8.0 N 
12.3 kg
 0.976 m/s 2
The sled is accelerating at 0.98 m/s2 in the positive x-direction.
y
x
b) Find the tension on the rope connecting the sleds.
Create the FBDs for each separate sled.
The tension on the rope between the sleds
doesn’t show up on the combined FBD.
The magnitude of the tension is the same for
each sled so we really on need one of the
single sled FBDs.
FNA
Tpull
FkA, TBA
FgA
v
sled A
Let’s look at sled B because it has fewer
|forces. Once again, we ignore the y-direction
forces so the forces are FkB and TAB .
y
The net force on sled B is
x
FNB
Fnet , x  TAB  FkB  mB ax
TAB
FkB
Rearrange to solve for the tension
TAB  mB ax  FkB

  5.6 kg  0.976 m/s
v
2
   8.0 N 
FgB
sled B
 13.46 N
The tension on the rope between the sleds is 13 N.
We have done this by starting with two FBDs, one for each sled. This would have generated two
equations with two unknowns a x and TAB . We could have then solved the two equations for the two
unknowns.
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
A related question is one where two crates are pushing on each other (instead of pulling in this case).
The force applied to the crates is also a push from behind the rear crate. The crates move at the same
acceleration so the acceleration can be found by treating the crates as one (you may have to look at
separate crates to find the friction forces). The force between the plates is Newton’s third law pair. It
can be found by looking at one crate by itself.
Another multiple crate problem has one crate stacked on the other. The force is from the side on one of
the crates. The acceleration can be found from the combined FBD and the friction is for the bottom
crate only. The friction force between the crates is a Newton’s third law pair. These questions usually
ask for the maximum or minimum coefficient of friction between the crates. It can usually be found
from the system acceleration and the FBD for the top crate.
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
PROBLEM 8
+y
A loonie (mL = 6.99 g) and a dime (mD = 2.09 g) are attached to the ends of a
thread. The thread lies over a smooth horizontal bar. Initially, the coins are
held motionless. When released, the coins start to move. The friction
between the thread and bar is negligible (zero).
+y
+y
a) Find the acceleration of the coins.
Draw FBDs
TLD
+y
choose directions
TDL
+y
FgD
FgL
Dime
Loonie
Note that the +y-direction is down for the loonie.
This is consistent with the direction definition of the dime. Assume that the thread doesn’t
stretch so downward motion of the loonie is the same as the upward motion of the dime.
We only have motion and forces in the y-direction.
The masses are defined in the question.
From the tension properties
TDL  TLD
or
TLD  TDL .
This is not true if the thread goes over a movable wheel or if there is friction.
Is the object accelerating? – yes, need to find a, use Newton’s second law.
The forces on the dime are the gravitational force and the tension, FgD  mD g and TLD . The net
force on the dime is
Fnet , D  FgD  TLD  mD a
All forces are in the same direction (±) so we can use scalars
mD g  TLD  mD a
Rearranging gives
TLD  mD a  mD g
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
The forces on the loonie are FgL  mL g and TDL  TLD .
The net force on the loonie is
Fnet , L  FgL  TDL  mLa
mL g  TLD  mL a
replace TDL with  TLD
mL g  TLD  mL a
all vectors in same direction, replace with scalars
Insert the expression for the tension TLD from the dime forces
mL g   mD a  mD g   mL a
More rearranging gives
mL a  mD a  mL g  mD g
a  mL  mD   g  mL  mD 
Isolating the acceleration
a
g  mL  mD 
 mL  mD 
9.80 m/s   6.99  2.09 

2
 6.99  2.09 
 5.289 m/s 2
The acceleration is 5.29 m/s2.
b) Find the tension in the thread.
We just sub our acceleration this back into one if the equations for the tension.
First convert the dime mass to kg.
mD = 2.09×10-3 kg
TLD  mD a  mD g


 

 2.09  103 kg 5.289 m/s 2 + 2.09  10 3 kg 9.80 m/s 2

 3.15  102 N
The tension on the tread is 3.1510-2 N.
There are cases where you have more pulley ropes. There are cases where the rope goes over a fixed
object or an object that moves. Check out examples in the assignments and text.
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PHYSICS SEMESTER ONE
UNIT 4: ANSWERS TO PROBLEMS
We could have treated the two masses as one, with the loonie weight as the force in one direction and
the dime weight as the force in the other. That would have given us the acceleration. We would then
use an individual coin FBD to find the tension in the tread.
Can you apply Newton’s Laws to a fixed object? No, this only works for objects that can undergo
acceleration.
NANSLO Physics Core Units and Laboratory Experiments
by the North American Network of Science Labs Online,
a collaboration between WICHE, CCCS, and BCcampus
is licensed under a Creative Commons Attribution 3.0 Unported License;
based on a work at rwsl.nic.bc.ca.
Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges.
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