Incline and Friction Examples
Physics 6A
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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:
Fkinetic  k  N
Fstatic  s  N
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!
Prepared by Vince Zaccone
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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
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.
y
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?
First we draw a diagram of the forces.
Normal
force
friction
x
Fpush
weight
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
First we draw a diagram of the forces.
Normal
force
friction
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.
x
Fpush
weight
Fstatic  s  N
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
First we draw a diagram of the forces.
Normal
force
friction
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.
x
Fpush
weight
Fstatic  s  N
The coefficient is given, but we need to find the normal force.
We can use the y-direction forces to find it.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
First we draw a diagram of the forces.
Normal
force
x
friction
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.
Fpush
weight
Fstatic  s  N
Fy  ma y
The coefficient is given, but we need to find the normal force.
We can use the y-direction forces to find it.
N  mg  0
Now we can calculate the maximum friction force.
N  mg


N  100kg  9.8 m2  980N
s
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
First we draw a diagram of the forces.
Normal
force
x
friction
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.
Fpush
weight
Fstatic  s  N
Fy  ma y
The coefficient is given, but we need to find the normal force.
We can use the y-direction forces to find it.
N  mg  0
Now we can calculate the maximum friction force.
N  mg
Fstatic  0.6   980N  588N
N  100kg  9.8 m2  980N

s

Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
First we draw a diagram of the forces.
Normal
force
x
friction
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.
Fpush
weight
Fstatic  s  N
Fy  ma y
The coefficient is given, but we need to find the normal force.
We can use the y-direction forces to find it.
N  mg  0
Now we can calculate the maximum friction force.
N  mg
Fstatic  0.6   980N  588N
N  100kg  9.8 m2  980N

s

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)?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
Normal
force
friction
x
Fpush
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:
Fx  ma x
Fpush  friction  ma x
weight
What type of friction do we have?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
Normal
force
friction
x
Fpush
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:
Fx  ma x
Fpush  friction  ma x
weight
The box is moving, so kinetic friction
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
Normal
force
friction
x
Fpush
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:
Fx  ma x
weight
Fpush  friction  ma x
Fpush  k  N  ma x
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
Normal
force
friction
x
Fpush
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:
Fx  ma x
weight
Fpush  friction  ma x
Fpush  k  N  ma x
588N  (0.5)  (980N)  (100kg)  a x
a x  0.98 m2
s
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
y
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?
Normal
force
friction
x
Fpush
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:
Fx  ma x
weight
Fpush  friction  ma x
Fpush  k  N  ma x
588N  (0.5)  (980N)  (100kg)  a x
a x  0.98 m2
s
Now that we have the acceleration we can use our kinematics formula:
x  v 0,x  t  21 a x  t 2
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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.
y
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?
Normal
force
friction
x
Fpush
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:
Fx  ma x
weight
Fpush  friction  ma x
Fpush  k  N  ma x
588N  (0.5)  (980N)  (100kg)  a x
a x  0.98 m2
s
Now that we have the acceleration we can use our kinematics formula:
x  v 0,x  t  1 a x  t 2
2
x  0  1  0.98 m2   3s 2
2
s 
x  4.41m
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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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Let’s do a more difficult version of this problem. This time the
push is not horizontal, but down at an angle.
Normal
force
friction
x
Fpush
Think about what will change if the push is not horizontal.
weight
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Let’s do a more difficult version of this problem. This time the
push is not horizontal, but down at an angle.
Normal
force
friction
x
Fpush
Think about what will change if the push is not horizontal.
Because the push is downward, the Normal force must
increase to compensate. This creates more friction than before.
weight
This might seem like a minor difference, but it makes the
algebra much more involved.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Let’s do a more difficult version of this problem. This time the
push is not horizontal, but down at an angle.
Normal
force
friction
x
Fpush
Think about what will change if the push is not horizontal.
Because the push is downward, the Normal force must
increase to compensate. This creates more friction than before.
weight
This might seem like a minor difference, but it makes the
algebra much more involved.
Before we can write the formulas, The pushing force needs to
be broken into components.
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For Campus Learning
Assistance Services at UCSB
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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Let’s do a more difficult version of this problem. This time the
push is not horizontal, but down at an angle.
Think about what will change if the push is not horizontal.
Because the push is downward, the Normal force must
increase to compensate. This creates more friction than before.
Normal
force
x
Fpush,x
friction
Fpush,y
weight
This might seem like a minor difference, but it makes the
algebra much more involved.
Before we can write the formulas, The pushing force needs to
be broken into components.
𝐹𝑝𝑢𝑠ℎ,𝑥 = 𝐹𝑝𝑢𝑠ℎ ∙ cos 30°
𝐹𝑝𝑢𝑠ℎ,𝑦 = 𝐹𝑝𝑢𝑠ℎ ∙ sin 30°
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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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Normal
force
Now we can write Newton’s 2nd law for each direction:
Fpush,x
friction
Σ𝐹𝑥 = 𝑚𝑎𝑥
x
Fpush,y
weight
Σ𝐹𝑦 = 𝑚𝑎𝑦
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Assistance Services at UCSB
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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Normal
force
Now we can write Newton’s 2nd law for each direction:
Fpush,x
friction
Σ𝐹𝑥 = 𝑚𝑎𝑥
x
Fpush,y
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝑚𝑎𝑥
weight
Σ𝐹𝑦 = 𝑚𝑎𝑦
𝑁 − 𝐹𝑝𝑢𝑠ℎ ∙ sin 30° − 𝑚𝑔 = 𝑚𝑎𝑦
As in the previous problem, we set the acceleration to 0 because the box is not going to
move unless we push harder than the maximum static friction.
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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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Normal
force
Now we can write Newton’s 2nd law for each direction:
Fpush,x
friction
Σ𝐹𝑥 = 𝑚𝑎𝑥
x
Fpush,y
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝑚𝑎𝑥
weight
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝜇𝑠 𝑁 = 0
Σ𝐹𝑦 = 𝑚𝑎𝑦
𝑁 − 𝐹𝑝𝑢𝑠ℎ ∙ sin 30° − 𝑚𝑔 = 𝑚𝑎𝑦
𝑁 − 𝐹𝑝𝑢𝑠ℎ ∙ sin 30° − 𝑚𝑔 = 0
At this point we have 2 equations with 2 unknowns (N and Fpush).
We can solve for N in the y-equation, then substitute into the x-equation.
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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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Normal
force
Now we can write Newton’s 2nd law for each direction:
Fpush,x
friction
Σ𝐹𝑥 = 𝑚𝑎𝑥
x
Fpush,y
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝑚𝑎𝑥
weight
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝜇𝑠 𝑁 = 0
Σ𝐹𝑦 = 𝑚𝑎𝑦
𝑁 − 𝐹𝑝𝑢𝑠ℎ ∙ sin 30° − 𝑚𝑔 = 𝑚𝑎𝑦
𝑁 − 𝐹𝑝𝑢𝑠ℎ ∙ sin 30° − 𝑚𝑔 = 0
𝑁 = 𝐹𝑝𝑢𝑠ℎ ∙ sin 30° + 𝑚𝑔
At this point we have 2 equations with 2 unknowns (N and Fpush).
We can solve for N in the y-equation, then substitute into the x-equation.
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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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Normal
force
Now we can write Newton’s 2nd law for each direction:
Fpush,x
friction
Σ𝐹𝑥 = 𝑚𝑎𝑥
x
Fpush,y
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝑚𝑎𝑥
weight
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝜇𝑠 𝑁 = 0
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝜇𝑠 (𝐹𝑝𝑢𝑠ℎ ∙ sin 30° + 𝑚𝑔) = 0
Now it is basic algebra to solve for Fpush.
Prepared by Vince Zaccone
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Assistance Services at UCSB
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 (at an angle of 30° below horizontal) until it
starts to slide. Once it starts to move, you keep pushing with constant force.
y
a)
How hard do you have to push to get the box moving?
Normal
force
Now we can write Newton’s 2nd law for each direction:
Fpush,x
friction
Σ𝐹𝑥 = 𝑚𝑎𝑥
x
Fpush,y
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝑚𝑎𝑥
weight
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝜇𝑠 𝑁 = 0
𝐹𝑝𝑢𝑠ℎ ∙ cos 30° − 𝜇𝑠 (𝐹𝑝𝑢𝑠ℎ ∙ sin 30° + 𝑚𝑔) = 0
Now it is basic algebra to solve for Fpush.
𝐹𝑝𝑢𝑠ℎ =
𝜇𝑠 𝑚𝑔
= 1039𝑁
cos 30° − 𝜇𝑠 ∙ sin 30°
As you can see, the extra downward component of the push makes it much harder to move the box.
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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)
θ
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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.
θ
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
Normal
FSisyphus
θ
W boulder,downhill
θ
W boulder
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
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:
Normal
FSisyphus
θ
W boulder,downhill
θ
W boulder
Wboulder ,downhill  FSisyphus
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
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:
Normal
FSisyphus
θ
W boulder,downhill
θ
W boulder
Wboulder ,downhill  FSisyphus
Wboulder   sin   FSisyphus
Wboulder 
FSisyphus
sin 
Now all we need is FSisyphus – How do we find that?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
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:
Normal
FSisyphus
θ
W boulder,downhill
θ
W boulder
Wboulder ,downhill  FSisyphus
Wboulder   sin   FSisyphus
Wboulder 
FSisyphus
sin 
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
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.
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:
Normal
FSisyphus
θ
W boulder,downhill
θ
W boulder
Wboulder ,downhill  FSisyphus
Wboulder   sin   FSisyphus
Wboulder 
FSisyphus
sin 
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.
Wboulder 
4900N
sin 20
 14,327N
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
Same as the last problem,
but now with added friction!
θ
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point?
θ
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point? DOWNHILL, opposing Sisyphus.W
Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away
from the incline. This way we can write down two force formulas.
y
x
Fnormal
FSisyphus
ƒstatic
θ
W boulder,y
boulder,x
θ
W boulder
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point? DOWNHILL, opposing Sisyphus.W
Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away
from the incline. This way we can write down two force formulas.
 Fx  ma x
 Fy  may
y
x
Fnormal
FSisyphus
ƒstatic
θ
W boulder,y
boulder,x
θ
W boulder
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point? DOWNHILL, opposing Sisyphus.W
Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away
from the incline. This way we can write down two force formulas.
 Fx  ma x
 Fy  may
y
x
Fnormal
FSisyphus
ƒstatic
θ
W boulder,y
boulder,x
θ
W boulder
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.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point? DOWNHILL, opposing Sisyphus.W
Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away
from the incline. This way we can write down two force formulas.
 Fx  0
 Fy  0
y
x
Fnormal
FSisyphus
ƒstatic
θ
W boulder,y
boulder,x
θ
W boulder
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point? DOWNHILL, opposing Sisyphus.W
Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away
from the incline. This way we can write down two force formulas.
 Fx  0
FSisyphus  fstatic  WBoulder ,x  0
 Fy  0
y
x
Fnormal
FSisyphus
ƒstatic
θ
W boulder,y
boulder,x
θ
W boulder
Fnormal  WBoulder ,y  0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point? DOWNHILL, opposing Sisyphus.W
Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away
from the incline. This way we can write down two force formulas.
 Fy  0
 Fx  0
FSisyphus  fstatic  WBoulder ,x  0
Fnormal  WBoulder ,y  0
4900N  s  Fnormal  WBoulder ,x  0
Fnormal  WBoulder ,y
y
x
Fnormal
FSisyphus
ƒstatic
θ
W boulder,y
boulder,x
θ
W boulder
We are assuming
Sisyphus can push
with a maximum
force of 4900 N.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point? DOWNHILL, opposing Sisyphus.W
y
x
Fnormal
FSisyphus
ƒstatic
θ
W boulder,y
boulder,x
Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away
from the incline. This way we can write down two force formulas.
 Fy  0
 Fx  0
FSisyphus  fstatic  WBoulder ,x  0
Fnormal  WBoulder ,y  0
4900N   s  Fnormal  WBoulder ,x  0
Fnormal  WBoulder  cos(20 )
θ
W boulder
4900N   s  Fnormal  WBoulder  sin(20 )  0
Here’s where we use our triangles
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point? DOWNHILL, opposing Sisyphus.W
y
x
Fnormal
FSisyphus
ƒstatic
θ
W boulder,y
boulder,x
Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away
from the incline. This way we can write down two force formulas.
 Fx  0
FSisyphus  fstatic  WBoulder ,x  0
Fnormal  WBoulder ,y  0
4900N  s  Fnormal  WBoulder ,x  0
Fnormal  WBoulder  cos(20 )
 Fy  0
θ
W boulder
4900N  s  Fnormal  WBoulder  sin(20 )  0


4900N  0.3   WBoulder  cos(20 )  WBoulder  sin(20 )  0
Now we can combine our equations by
substituting for Fnormal in the x equation.
We also know the coefficient of friction.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
This time we will have to include a friction force in our diagram.
Which direction should it point? DOWNHILL, opposing Sisyphus.W
y
x
Fnormal
FSisyphus
ƒstatic
θ
W boulder,y
boulder,x
Notice that this time we have labeled an axis system with the xdirection pointing uphill, and the y-direction pointing directly away
from the incline. This way we can write down two force formulas.
 Fx  0
FSisyphus  fstatic  WBoulder ,x  0
Fnormal  WBoulder ,y  0
4900N   s  Fnormal  WBoulder ,x  0
Fnormal  WBoulder  cos(20 )
 Fy  0
θ
W boulder
4900N   s  Fnormal  WBoulder  sin(20 )  0


4900N  0.3   WBoulder  cos(20 )  WBoulder  sin(20 )  0
WBoulder  7850N
The final step is just a bit of algebra.
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