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Incline and Friction Examples Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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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

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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 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

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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 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 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!

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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 µ k =0.5 and µ s =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

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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 µ k =0.5 and µ s =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?

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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. 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 µ k =0.5 and µ s =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?

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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. 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 µ k =0.5 and µ s =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?

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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. 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 Example – pushing a box across the floor There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µ k =0.5 and µ s =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?

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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. 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 Example – pushing a box across the floor There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µ k =0.5 and µ s =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?

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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. 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 This is how hard we have to push to get the box moving (ok, maybe we push with a force of N) 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 µ k =0.5 and µ s =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?

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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 2 nd law for the x-direction: 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 µ k =0.5 and µ s =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?

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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 2 nd law for the x-direction: 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 µ k =0.5 and µ s =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?

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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 2 nd law for the x-direction: Example – pushing a box across the floor There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µ k =0.5 and µ s =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?

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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 2 nd law for the x-direction: Example – pushing a box across the floor There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µ k =0.5 and µ s =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?

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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 2 nd law for the x-direction: Now that we have the acceleration we can use our kinematics formula: Example – pushing a box across the floor There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µ k =0.5 and µ s =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?

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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 2 nd law for the x-direction: Now that we have the acceleration we can use our kinematics formula: Example – pushing a box across the floor There is a 100kg box at rest on a horizontal floor. The coefficients of friction are µ k =0.5 and µ s =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?

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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

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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

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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. F Sisyphus W boulder,downhill θ W boulder Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Normal

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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. F Sisyphus W boulder,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: W boulder Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Normal

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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. F Sisyphus W boulder,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: W boulder Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Now all we need is F Sisyphus – How do we find that? Normal

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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. F Sisyphus W boulder,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: W 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

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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. F Sisyphus W boulder,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: W 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

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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!

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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?

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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. F Sisyphus W boulder,x θ W boulder ƒ static W boulder,y y x F normal

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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. F Sisyphus W boulder,x θ W boulder ƒ static W boulder,y y x F normal

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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. F Sisyphus W boulder,x θ W boulder ƒ static W boulder,y y x F normal 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.

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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. F Sisyphus W boulder,x θ W boulder ƒ static W boulder,y y x F normal

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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. F Sisyphus W boulder,x θ W boulder ƒ static W boulder,y y x F normal

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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. F Sisyphus W boulder,x θ W boulder ƒ static W boulder,y y x F normal We are assuming Sisyphus can push with a maximum force of 4900 N.

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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. F Sisyphus W boulder,x θ W boulder ƒ static W boulder,y y x F normal Here’s where we use our triangles

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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. F Sisyphus W boulder,x θ W boulder ƒ static W boulder,y y x F normal Now we can combine our equations by substituting for F normal in the x equation. We also know the coefficient of friction.

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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. F Sisyphus W boulder,x θ W boulder ƒ static W boulder,y y x F normal The final step is just a bit of algebra.

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