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Published byCaiden Polley Modified about 1 year ago

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Friction

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Friction - the force needed to drag one object across another. (at a constant velocity) Depends on: How hard the surfaces are held together What type of surface it is (i.e. rough, smooth) Not supposed to depend on: Surface area (pressure) Speed (low speeds)

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F Fr = μF N Force of Friction in N Coefficient of Friction. 0 < μ < 1 Normal Force - Force exerted by a surface to maintain its integrity Usually the weight (level surfaces)

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K inetic Friction - Force needed to k eep it going at a constant velocity. F Fr = μ k F N Always in opposition to velocity St atic Friction - Force needed to st art motion. F Fr < μ s F N Keeps the object from moving if it can. Only relevant when object is stationary. Always in opposition to applied force. Calculated value is a maximum

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A 5.0 kg block of wood has a μ s of 0.41, and a μ k of.24 between itself and the level floor. a. Calculate the limit of static friction, and the kinetic friction ≈ 20. N, ≈ 12 N 5.0 kg

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A 5.0 kg block of wood has a μ s of 0.41, and a μ k of.24 between itself and the level floor. ( N, N) b. The block is at rest, and you exert a force of 15 N to the right to try to make it move. Draw and label all the forces acting on the block 5.0 kg

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A 5.0 kg block of wood has a μ s of 0.41, and a μ k of.24 between itself and the level floor. ( N, N) c. If the block is initially at rest, and we exert a force of 35 N to the right, calculate the block’s acceleration ≈ 4.6 m/s/s 5.0 kg

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A 5.0 kg block of wood has a μ s of 0.41, and a μ k of.24 between itself and the level floor. ( N, N) d. Kyle and Sally both exert a force on the block, and it accelerates from rest to a 5.2 m/s over a distance of 10. m to the right. If Kyle exerts a force of 25 N to the right, what force in what direction does Sally exert? ≈ -6.5 N 5.0 kg

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Whiteboards: Friction 11 | 2 | 3 | 4 | 5 | TOC

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What is the force needed to drag a 12 kg chunk of rubber at a constant velocity across dry concrete? F = ma, F Fr = μ k F N F N = weight = mg = (12 kg)(9.8 N/kg) = N F Fr = μ k F N = (.8)(117.6 N) = N = 90 N 90 N

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What is the force needed to start a 150 kg cart sliding across wet concrete from rest if the wheels are locked up? F = ma, F Fr < μ s F N F N = weight = mg = (150 kg)(9.8 N/kg) = 1470 N F Fr = μ k F N = (.7)(1470) = 1029 N = 1000 N 1000 N

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μ s =.62, μ k =.48 What is the acceleration if there is a force of 72 N in the direction an 8.5 kg block is already sliding? F Fr = μ k F N, F N = mg, F Fr = μ k mg F Fr = (.48)(8.5 kg)(9.8 N/kg) = N F = ma = (8.5 kg)a, a = 3.77 = 3.8 ms m/s/s 72 N 8.5 kg v F Fr

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μ s =.62, μ k =.48 A 22 kg block is sliding on a level surface initially at 12 m/s. What time to stop? F Fr = μ k F N, F N = mg, F Fr = μ k mg F Fr = (.48)(22 kg)(9.8 N/kg) = N F = ma = (22 kg)a, a = ms -2 v = u + at, v = 0, u = 12, a = ms -2, t = 2.55 s = 2.6 s 2.6 s 22 kg v=12m/s F Fr

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μ s =.62, μ k =.48 A 6.5 kg box accelerates and moves to the right at 3.2 m/s/s, what force must be applied? F Fr = μ k F N, m = 6.5 kg F Fr = (.48)(6.5 kg)(9.8 N/kg) = N F = ma = (6.5 kg)(3.2 ms -2 ), F = = 51 N 51 N F = ? F Fr 6.5 kg v a = 3.2 ms -2

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μ s =.62, μ k =.48 A 22 kg block is sliding on a level surface initially at 12 m/s stops in 2.1 seconds. What external force is acting on it besides friction?? v = u + at, v = 0, u = 12 m/s, t = 2.1 s, a = ms -2 F Fr = μ k F N, F N = mg, F Fr = μ k mg F Fr = (.48)(22 kg)(9.8 N/kg) = N F = ma = (22 kg)( ms -2 ), F = -22 N (left) -22 N (to the left) 22 kg v=12m/s F Fr F = ?

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μ s =.62, μ k =.48 A force of 35 N in the direction of motion accelerates a block at 1.2 m/s/s in the same direction What is the mass of the block? F Fr = μ k F N, F N = mg, F Fr = μ k mg F Fr = (.48)m(9.8 ms -2 ) = m(4.704 ms -2 ) F = ma = m(1.2 ms -2 ) 35 N = m(4.704 ms -2 ) + m(1.2 ms -2 ) = m(4.704 ms ms -2 ) m = (35 N)/(5.904 ms -2 ) = kg = 5.9 kg 5.9 kg 35 N F Fr m v a = 1.2 ms -2

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