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Incline and Friction Examples
Physics 6A
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Friction is a force that opposes the motion of surfaces that are in contact with each other.
We will consider 2 types of friction in this class:
KINETIC Friction – for surfaces that are in motion (sliding)
STATIC Friction – for surfaces at rest
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Friction is a force that opposes the motion of surfaces that are in contact with each other.
We will consider 2 types of friction in this class:
KINETIC Friction – for surfaces that are in motion (sliding)
STATIC Friction – for surfaces at rest
The formulas are very similar – each one has a “coefficient of friction” (µ) that determines how much of the Normal force is translated into friction force.
Crucial distinction – kinetic friction will be a constant force, while static friction will be just strong enough to keep the surfaces from slipping
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Friction is a force that opposes the motion of surfaces that are in contact with each other.
We will consider 2 types of friction in this class:
KINETIC Friction – for surfaces that are in motion (sliding)
STATIC Friction – for surfaces at rest
The formulas are very similar – each one has a “coefficient of friction” (µ) that determines how much of the Normal force is translated into friction force.
Crucial distinction – kinetic friction will be a constant force, while static friction will be just strong enough to keep the surfaces from slipping
Here are the formulas:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
NF
NF
sstatic
kkinetic
See – friction is FUN!
Static friction will have a maximum value. If you push any harder the surfaces will slip and you get kinetic friction instead!
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
First we draw a diagram of the forces.
weight
Normal force
Fpushfriction
x
y
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
First we draw a diagram of the forces.
weight
Normal force
Fpushfriction
To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction.
NF sstatic
x
y
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
First we draw a diagram of the forces.
weight
Normal force
Fpushfriction
To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction.
NF sstatic
The coefficient is given, but we need to find the normal force. We can use the y-direction forces to find it.
x
y
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
First we draw a diagram of the forces.
weight
Normal force
Fpushfriction
To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction.
NF sstatic
The coefficient is given, but we need to find the normal force. We can use the y-direction forces to find it.
Now we can calculate the maximum friction force.
x
y
N9808.9kg100N
mgN
0mgN
maF
2sm
yy
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
First we draw a diagram of the forces.
weight
Normal force
Fpushfriction
To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction.
NF sstatic
The coefficient is given, but we need to find the normal force. We can use the y-direction forces to find it.
Now we can calculate the maximum friction force.
x
y
N9808.9kg100N
mgN
0mgN
maF
2sm
yy
N588N9806.0Fstatic
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
First we draw a diagram of the forces.
weight
Normal force
Fpushfriction
To get the box moving we have to push hard enough to overcome static friction. So we need to find the maximum force of static friction.
NF sstatic
The coefficient is given, but we need to find the normal force. We can use the y-direction forces to find it.
Now we can calculate the maximum friction force.
x
y
N9808.9kg100N
mgN
0mgN
maF
2sm
yy
N588N9806.0Fstatic
This is how hard we have to push to get the box moving (ok, maybe we push with a force of 588.0000001N)
Now that we have the answer for part a) how do we do part b)?
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
weight
Normal force
Fpushfriction
x
y
Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas.
We can write down Newton’s 2nd law for the x-direction:
xpush
xx
mafrictionF
maF
What type of friction do we have?
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
weight
Normal force
Fpushfriction
x
y
Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas.
We can write down Newton’s 2nd law for the x-direction:
xpush
xx
mafrictionF
maF
The box is moving, so kinetic friction
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
weight
Normal force
Fpushfriction
x
y
Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas.
We can write down Newton’s 2nd law for the x-direction:
xkpush
xpush
xx
maNF
mafrictionF
maF
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
weight
Normal force
Fpushfriction
x
y
Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas.
We can write down Newton’s 2nd law for the x-direction:
2sm
x
x
xkpush
xpush
xx
98.0a
a)kg100()N980()5.0(N588
maNF
mafrictionF
maF
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
weight
Normal force
Fpushfriction
x
y
Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas.
We can write down Newton’s 2nd law for the x-direction:
Now that we have the acceleration we can use our kinematics formula:2
x21
x,0 tatvx
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
2sm
x
x
xkpush
xpush
xx
98.0a
a)kg100()N980()5.0(N588
maNF
mafrictionF
maF
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
weight
Normal force
Fpushfriction
x
y
Part b) is really a kinematics problem. To find the answer we need the acceleration of the box. Then we can go back to chapter 2 and use one of our kinematics formulas.
We can write down Newton’s 2nd law for the x-direction:
Now that we have the acceleration we can use our kinematics formula:
m41.4x
s398.00x
tatvx
2
sm
21
2x2
1x,0
2
Example – pushing a box across the floor
There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µk=0.5 and µs=0.6. You want to push it across the room, so you push (horizontally) until it starts to slide. Once it starts to move, you keep pushing with constant force.
a) How hard do you have to push to get the box moving?
b) How far will the box travel if you push for 3 seconds?
2sm
x
x
xkpush
xpush
xx
98.0a
a)kg100()N980()5.0(N588
maNF
mafrictionF
maF
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction)
We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction)
We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components.
FSisyphus
Wboulder,downhill θ
Wboulder
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Normal
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction)
We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components.
FSisyphus
Wboulder,downhill θ
For Sisyphus to be able to push the boulder up the hill, his force must be at least equal to the downhill component of the boulder’s weight. We can write down the formula like this: Wboulder
Sisyphusdownhill,boulder FW
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Normal
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction)
We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components.
FSisyphus
Wboulder,downhill θ
For Sisyphus to be able to push the boulder up the hill, his force must be at least equal to the downhill component of the boulder’s weight. We can write down the formula like this: Wboulder
sin
FW
FsinW
FW
Sisyphusboulder
Sisyphusboulder
Sisyphusdownhill,boulder
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Now all we need is FSisyphus – How do we find that?
Normal
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction)
We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components.
FSisyphus
Wboulder,downhill θ
For Sisyphus to be able to push the boulder up the hill, his force must be at least equal to the downhill component of the boulder’s weight. We can write down the formula like this: Wboulder
sin
FW
FsinW
FW
Sisyphusboulder
Sisyphusboulder
Sisyphusdownhill,boulder
Sisyphus’ force can be found from the given information. He can lift 500kg, so multiplying by g, his force is 4900N. The angle is given as 20°, so we can plug in to find our answer.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Normal
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. (ignore friction)
We can draw a diagram of forces for this problem. Sisyphus will push up the incline, but the weight of the boulder will be straight down, so we will split that one into components.
FSisyphus
Wboulder,downhill θ
For Sisyphus to be able to push the boulder up the hill, his force must be at least equal to the downhill component of the boulder’s weight. We can write down the formula like this: Wboulder
sin
FW
FsinW
FW
Sisyphusboulder
Sisyphusboulder
Sisyphusdownhill,boulder
Sisyphus’ force can be found from the given information. He can lift 500kg, so multiplying by g, his force is 4900N. The angle is given as 20°, so we can plug in to find our answer.
N327,1420sin
N4900Wboulder
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Normal
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Same as the last problem, but now with added friction!
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point?
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus.
Notice that this time we have labeled an axis system with the x-direction pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas.
FSisyphus
Wboulder,x
θ
Wboulder
ƒstatic
Wboulder,y
y xFnormal
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus.
Notice that this time we have labeled an axis system with the x-direction pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas.
FSisyphus
Wboulder,x
θ
Wboulder
ƒstatic
Wboulder,y
y xFnormal
xx maF yy maF
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus.
Notice that this time we have labeled an axis system with the x-direction pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas.
FSisyphus
Wboulder,x
θ
Wboulder
ƒstatic
Wboulder,y
y xFnormal
xx maF yy maF
In this type of problem we need all the forces to balance out. Even though we want Sisyphus to be able to lift the boulder, we want to be just on the borderline between when the boulder moves and when it doesn’t. Thus we want to be in equilibrium to
find the maximum weight. Equilibrium means zero acceleration.
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus.
Notice that this time we have labeled an axis system with the x-direction pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas.
FSisyphus
Wboulder,x
θ
Wboulder
ƒstatic
Wboulder,y
y xFnormal
0Fx 0Fy
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus.
Notice that this time we have labeled an axis system with the x-direction pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas.
FSisyphus
Wboulder,x
θ
Wboulder
ƒstatic
Wboulder,y
y xFnormal
0WfF
0F
x,BoulderstaticSisyphus
x
0WF
0F
y,Bouldernormal
y
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus.
Notice that this time we have labeled an axis system with the x-direction pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas.
FSisyphus
Wboulder,x
θ
Wboulder
ƒstatic
Wboulder,y
y xFnormal
0WFN4900
0WfF
0F
x,Bouldernormals
x,BoulderstaticSisyphus
x
y,Bouldernormal
y,Bouldernormal
y
WF
0WF
0F
We are assuming Sisyphus can push with a maximum force of 4900 N.
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus.
Notice that this time we have labeled an axis system with the x-direction pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas.
FSisyphus
Wboulder,x
θ
Wboulder
ƒstatic
Wboulder,y
y xFnormal
0)20sin(WFN4900
0WFN4900
0WfF
0F
Bouldernormals
x,Bouldernormals
x,BoulderstaticSisyphus
x
)20cos(WF
0WF
0F
Bouldernormal
y,Bouldernormal
y
Here’s where we use our triangles
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus.
Notice that this time we have labeled an axis system with the x-direction pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas.
FSisyphus
Wboulder,x
θ
Wboulder
ƒstatic
Wboulder,y
y xFnormal
0)20sin(W)20cos(W3.0N4900
0)20sin(WFN4900
0WFN4900
0WfF
0F
BoulderBoulder
Bouldernormals
x,Bouldernormals
x,BoulderstaticSisyphus
x
)20cos(WF
0WF
0F
Bouldernormal
y,Bouldernormal
y
Now we can combine our equations by substituting for Fnormal in the x equation.
We also know the coefficient of friction.
θ
Sisyphus is attempting to push this giant boulder up an incline. Using all his strength he can lift 500kg straight up over his head. If the incline is 20°, find the maximum weight (in Newtons) that Sisyphus can push up the hill. Assume the coefficient of static friction is 0.3.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
This time we will have to include a friction force in our diagram. Which direction should it point? DOWNHILL, opposing Sisyphus.
Notice that this time we have labeled an axis system with the x-direction pointing uphill, and the y-direction pointing directly away from the incline. This way we can write down two force formulas.
FSisyphus
Wboulder,x
θ
Wboulder
ƒstatic
Wboulder,y
y xFnormal
N7850W
0)20sin(W)20cos(W3.0N4900
0)20sin(WFN4900
0WFN4900
0WfF
0F
Boulder
BoulderBoulder
Bouldernormals
x,Bouldernormals
x,BoulderstaticSisyphus
x
)20cos(WF
0WF
0F
Bouldernormal
y,Bouldernormal
y
The final step is just a bit of algebra.
