Download Newtons Lesson 7

NEWTON’S LESSON 7
Finding Individual Forces
If the numerical value for the net force and the direction of the net force is known, then
the value of all individual forces can be determined.
Example:
Free-body diagrams for four situations are shown below. The net force is known for each
situation. However, the magnitudes of a few of the individual forces are not known.
Analyze each situation individually and determine the magnitude of the unknown forces.
Answer: A = 50 N (the horizontal forces must be balanced)
B = 200 N (the vertical forces must be balanced)
C = 1100 N (in order to have a net force of 200 N, up)
D = 20 N (in order to have a net force of 60 N, left)
E = 300 N (the vertical forces must be balanced)
F = H = any number you wish (as long as F equals H)
G = 50 N (in order to have a net force of 30 N, right)
Example:
A rightward force is applied to a 6-kg object to move it across a rough surface at constant
velocity. The object encounters 15 N of frictional force. Use the diagram to determine the
gravitational force, normal force, net force, and applied force. (Neglect air resistance.)
Answer: Fnet = 0 N; Fgrav = 58.8 N; Fnorm = 58.8 N; Fapp = 15 N
When the velocity is constant, a = 0 m/s/s and Fnet = 0 N
Since the mass is known, Fgrav can be found: Fgrav = m • g = 6 kg • 9.8 m/s/s = 58.8 N
Since there is no vertical acceleration, the normal force equals the gravity force.
Since there is no horizontal acceleration, Ffrict = Fapp = 15 N
Example:
A rightward force is applied to a 5-kg object to move it across a rough surface with a
rightward acceleration of 2 m/s/s. The coefficient of friction between the object and the
surface is 0.1. Use the diagram to determine the gravitational force, normal force, applied
force, frictional force, and net force. (Neglect air resistance.)
Answer: Fnet =10 N, right; Fgrav = 49 N; Fnorm = 49 N; Ffrict = 4.9 N; Fapp = 14.9 N
Fnet can be found using Fnet = m • a = (5 kg) • (2 m/s/s) = 10 N, right.
Since the mass is known, Fgrav can be found: Fgrav = m • g = 5 kg • 9.8 m/s/s = 49 N
Since there is no vertical acceleration, the normal force equals the gravity force.
Once Fnorm is known, Ffrict can be found using Ffrict = "mu" • Fnorm = (0.1) • (49 N) = 4.9
N.
Since the Fnet = 10 N, right, the rightward force (Fapp) must be 10 N more than the
leftward force (Ffrict); thus, Fapp must be 14.9 N.
Forces in Two Dimensions
Resolution of Forces
What if a force vector isn’t horizontal or vertical? How do you add a force vector that is
at an angle?
Any vector which is directed at an angle to the x-y axis can be considered to have two
parts - each part being directed along one of the axes - either horizontally or
vertically.
EXAMPLE: Rex is being pulled upon by a dog chain. If the chain is pulled
upwards and to the right.
That single force can be resolved into two components - one directed upwards
and the other directed rightwards.
Each component describes the influence of that chain in the given direction.
The vertical component describes the upward influence of the force upon Rex and the
horizontal component describes the rightward influence of the force upon Rex.
Assume that the chain is exerting a 60 N force upon Rex at an angle of 40 degrees above
the horizontal.
EXAMPLE: A 400-N force is exerted at a 60-degree angle to move a railroad car
eastward along a
railroad track.
What happens when the angle which a force makes with the horizontal increase?
The component of force in the horizontal direction (Fx) decreases. The more that a force
is directed upwards (the angle with the horizontal increases), the less that the force is able
to exert an influence in the horizontal direction.
Therefore if you wish to drag Fido horizontally, what would be the best angle above
the horizontal would you want to pull him at?
You would make an effort to pull in as close to a horizontal direction as possible.
What if you pulled vertically on Rex’s chain, would he move horizontally?
No, since there is no horizontal component to a vertical force, therefore you probably
would just lift Rex off the ground.
One important application of this principle is in the recreational sport of sail boating.
Sailboats encounter a force of wind resistance due to the impact of the moving air
molecules against the sail. The force of wind resistance is directed perpendicular to the
face of the sail, and as such is often directed at an angle to the direction of the sailboat's
motion.
To determine the influence of the wind resistance force in the direction of motion, that
force will have to be resolved into two components - one in the direction which the
sailboat is moving and the other in a direction perpendicular to the sailboat's motion.
Which case would produce the greatest influence in the direction of the sailboat's motion?
That is, which case has the greatest component of force in the direction parallel to the
boats' heading?
Many people believe that a sailboat cannot travel "upwind."
It is their perception that if the wind blows from north to south, then there is no possible
way for a sailboat to travel from south to north.
This is simply not true. Sailboats can travel "upwind" and commonly do so by a method
known as tacking into the wind.
It is true to say that a sailboat can never travel upwind by heading its boat
directly into the wind.
If the boat heads directly into the wind, then the wind force is directed due
opposite its heading. In such a case, there is no component of force in the
direction that the sailboat is heading. That is, there is no "propelling force."
On the other hand, if the boat heads at an angle into the wind, then the wind
force can be resolved into two components.
NEWTON’S LESSON 7 HOMEWORK
1. A rightward force is applied to a 10-kg object to move it across a rough surface at
constant velocity. The coefficient of friction between the object and the surface is 0.2.
Use the diagram to determine the gravitational force, normal force, applied force,
frictional force, and net force. (Neglect air resistance.)
2. Lee Mealone is sledding with his friends when he becomes disgruntled by one of his
friends comments. He exerts a rightward force of 9.13 N on his 4.68-kg sled to accelerate
it across the snow. If the acceleration of the sled is 0.815 m/s/s, determine the
gravitational force, normal force, frictional force, net force, and the coefficient of friction
between the object and the surface. (Neglect air resistance.)
3. In a Physics lab, Ernesto and Amanda apply a 34.5 N rightward force to a 4.52-kg cart
to accelerate it across a horizontal surface at a rate of 1.28 m/s/s. Determine the friction
force acting upon the cart.
4. A rightward force of 25 N is applied to a 4-kg object to move it across a rough surface
with a rightward acceleration of 2.5 m/s/s. Use the diagram to determine the gravitational
force, normal force, frictional force, net force, and the coefficient of friction between the
object and the surface. (Neglect air resistance.)
5. The diagram at the right depicts a force which makes an angle to the horizontal.
This force will have horizontal and vertical components. Which one of the choices
below best depicts the direction of the horizontal and vertical components of this
force?
6. Determine the components of each of the following vectors
7. Three sailboats are shown below. Each sailboat experiences the same amount of force,
yet has different sail orientations.
In which case (A, B or C) is the sailboat most likely to tip over sideways? Explain.
8. Consider the tow truck at the right. If the tensional
force in the cable is 1000 N and if the cable makes a 60degree angle with the horizontal, then what is the vertical
component of force which lifts the car off the ground?
9. A rescue plane wants to drop supplies to isolated
mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally
with a speed of 250.0 km/h, and the plane releases the supplies a horizontal distance of
425 m in advance of the mountain climbers, what vertical velocity (up or down) should
the supplies be given so that they arrive precisely at the climbers’ position? THIS IS A
THINKING QUESTION! Draw a diagram and think before solving.
HOMEWORK KEY
1.
2.
3.
4.
5.
6.
7.
8.
9.
Fnet = 0 N; Fgrav = 98 N; Fnorm = 98 N; Ffrict = 19.6 N; Fapp = 19.6 N
Fgrav = 45.9 N; Fnorm = 45.9 N; Ffrict = 3.07 N; Fnet = 3.81 N, right; μ = 0.116
Ffrict = 28.7 N, left
Fnet = 10 N, right; Fgrav = 39.2 N; Fnorm = 39.2 N; Ffrict = 15 N; μ=0.383
D
A: Fx 86.6 N, Fy 50.0 N B: Fx 70.7 N, Fy 70.7 N C: Fx 50.0 N, Fy 86.6 N
A
866 N, upward.
8.41 m/s down
NEWTON’S LESSON 7 HOMEWORK
1. A rightward force is applied to a 10-kg object to move it across a rough surface at
constant velocity. The coefficient of friction between the object and the surface is 0.2.
Use the diagram to determine the gravitational force, normal force, applied force,
frictional force, and net force. (Neglect air resistance.)
Answer: Fnet = 0 N; Fgrav = 98 N; Fnorm = 98 N; Ffrict = 19.6 N; Fapp = 19.6 N
When the velocity is constant, a = 0 m/s/s and Fnet = 0 N
Since the mass is known, Fgrav can be found (Fgrav = m • g = 10 kg • 9.8 m/s/s). Since
there is no vertical acceleration, normal force = gravity force. Once Fnorm is known, Ffrict
can be found using Ffrict = "mu" •Fnorm. Since there is no horizontal acceleration, Ffrict =
Fapp = 19.6 N
2. Lee Mealone is sledding with his friends when he becomes disgruntled by one of his
friends comments. He exerts a rightward force of 9.13 N on his 4.68-kg sled to accelerate
it across the snow. If the acceleration of the sled is 0.815 m/s/s, determine the
gravitational force, normal force, frictional force, net force, and the coefficient of friction
between the object and the surface. (Neglect air resistance.)
Answer:
Fgrav = 45.9 N; Fnorm = 45.9 N; Ffrict = 3.07 N; Fnet = 3.81 N, right; μ = 0.116
The starting point for any problem such as this is the construction of a
free-body diagram in which you show all the individual forces which are
acting upon the book. There are two vertical forces - gravity and normal
force. There are two horizontal forces - friction and the applied force.
Since there is no vertical acceleration, normal force = gravity force. Each
of these forces can be determined using the equation Fgrav = m • g = (4.68
kg) • (9.8 m/s/s) = 45.864 N
The net force can be determined from knowledge of the mass and
acceleration of the sled. Fnet = m • a = (4.68 kg) • (0.815 m/s/s) = 3.8142
N, right.
Since the net force is to the right (in the direction of the applied force),
then the applied force must be greater than the friction force. The friction
force can be determined using an understanding of net force as the vector
sum of all the forces. So 3.81 N, right = 9.13 N, right + Ffrict. Therefore,
Ffrict = 5.32 N, left.
The coefficient of friction can now be determined as the ratio of friction
force to normal force. μ = Ffrict / Fnorm = (5.32 N) / (45.864 N) = 0.116.
3. In a Physics lab, Ernesto and Amanda apply a 34.5 N rightward force to a 4.52-kg cart
to accelerate it across a horizontal surface at a rate of 1.28 m/s/s. Determine the friction
force acting upon the cart.
Answer:
Ffrict = 28.7 N, left
The starting point for any problem such as this is the construction of a
free-body diagram in which you show all the individual forces which are
acting upon the book. There are two vertical forces - gravity and normal
force. There are two horizontal forces - friction and the applied force.
Since there is no vertical acceleration, normal force = gravity force. Each
of these forces can be determined using the equation Fgrav = m • g = (4.525
kg) • (9.8 m/s/s) = 44.296 N
The net force can be determined from knowledge of the mass and
acceleration of the sled. Fnet = m • a = (4.52 kg) • (1.28 m/s/s) = 5.7856 N,
right.
Since the net force is to the right (in the direction of the applied force),
then the applied force must be greater than the friction force. The friction
force can be determined using an understanding of net force as the vector
sum of all the forces. So 5.7856 N, right = 34.5 N, right + Ffrict. Therefore,
Ffrict = 28.7 N, left.
4. A rightward force of 25 N is applied to a 4-kg object to move it across a rough surface
with a rightward acceleration of 2.5 m/s/s. Use the diagram to determine the gravitational
force, normal force, frictional force, net force, and the coefficient of friction between the
object and the surface. (Neglect air resistance.)
Fnet = 10 N, right; Fgrav = 39.2 N; Fnorm = 39.2 N; Ffrict = 15 N; μ=0.383
Fnet can be found using Fnet = m • a = (4 kg) • (2.5 m/s/s) =10 N, right.
Since the mass is known, Fgrav can be found: Fgrav = m • g = 4 kg • 9.8 m/s/s = 39.2 N.
Since there is no vertical acceleration, the normal force equals the gravity force.
Since the Fnet=10 N, right, the rightward force (Fapp) must be 10 N more than the leftward
force (Ffrict); thus, Ffrict must be 15 N.
Finally, µ = Ffrict / Fnorm = (15 N) / (39.2 N) = 0.383.
5. The diagram at the right depicts a force which makes an angle to the horizontal.
This force will have horizontal and vertical components. Which one of the choices
below best depicts the direction of the horizontal and vertical components of this
force?
The answer is D.
The force is directed downwards and leftwards. Thus, this force will have a downward
vertical component and a leftward horizontal component. This would be consistent with
choice d.
6. Determine the components of each of the following vectors
Answer:
A: cos (30 degrees) = Fx / (100N)
Fx = 100 N • cos (30 degrees) = 86.6 N
sin (30 degrees) = (Fy / (100N)
Fy = 100 N * sin (30 degrees) = 50.0 N
B: cos (45 degrees) = Fx / (100N)
Fx = 100 N • cos (45 degrees) = 70.7 N
sin (45 degrees) = Fy / (100N)
Fy = 100 N • sin (45 degrees) = 70.7 N
C: cos (60 degrees) = Fx / (100N)
Fx = 100 N • cos (60 degrees) = 50.0 N
sin (60 degrees) = Fy/ (100N)
Fy = 100 N • sin (60 degrees) = 86.6 N
7. Three sailboats are shown below. Each sailboat experiences the same amount of force,
yet has different sail orientations.
In which case (A, B or C) is the sailboat most likely to tip over sideways? Explain.
The answer is Case A.
While it is the parallel component of force which propels the boat forward, it is the
perpendicular component of force which tips the boat over. This component of force is
greatest in Case A as seen in the diagram.
8. Consider the tow truck at the right. If the tensional
force in the cable is 1000 N and if the cable makes a 60degree angle with the horizontal, then what is the vertical
component of force which lifts the car off the ground?
The answer is 866 N, upward.
The vertical component can be found if a triangle is constructed with the 1000 N diagonal
force as the hypotenuse. The vertical component is the length of the side opposite the
hypotenuse. Thus,
sin (60 degrees) = (Fvert) / (1000 N)
Solving for Fvert will give the answer 866 N.
9. A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge
235 m below. If the plane is traveling horizontally with a speed of 250.0 km/h, and
the plane releases the supplies a horizontal distance of 425 m in advance of the
mountain climbers, what vertical velocity (up or down) should the supplies be given
so that they arrive precisely at the climbers’ position? THIS IS A THINKING
QUESTION! Draw a diagram and think before solving.
Answer: If the goods are given a vertical velocity, they will still have the horizontal
velocity of the plane. We find the time to fall from:
250 km/h  69.4444 m/s
dxf = vxi*t
425 m = (69.4444m/s)*t
t = 6.1200 s
We find the necessary vertical velocity from
dfy – diy= viyt + ½ at2
0-235 = viy(6.12) + ½ (-9.8)(6.12)2
viy = -8.41 m/s = 8.41 m/s down