Friction

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)

FFr = μFNForce 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)

Kinetic Friction - Force needed to keep it going at a constant velocity.

FFr = μkFN

Always in opposition to velocity

Static Friction - Force needed to start motion.FFr < μsFN

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

A 5.0 kg block of wood has a μs of 0.41, and a μk of .24 between itself and the level floor. a. Calculate the limit of static friction, and the kinetic friction.

20.1105 ≈ 20. N, 11.772 ≈ 12 N

5.0 kg5.0 kg

A 5.0 kg block of wood has a μs of 0.41, and a μk of .24 between itself and the level floor. (20.1105 N, 11.772 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 kg5.0 kg

A 5.0 kg block of wood has a μs of 0.41, and a μk of .24 between itself and the level floor. (20.1105 N, 11.772 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.

4.6456 ≈ 4.6 m/s/s

5.0 kg5.0 kg

A 5.0 kg block of wood has a μs of 0.41, and a μk of .24 between itself and the level floor. (20.1105 N, 11.772 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?

-6.468 ≈ -6.5 N

5.0 kg5.0 kg

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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, FFr = μkFN

FN = weight = mg = (12 kg)(9.8 N/kg) = 117.6 NFFr = μkFN = (.8)(117.6 N) = 94.08 N = 90 N

90 N

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, FFr < μsFN

FN = weight = mg = (150 kg)(9.8 N/kg) = 1470 NFFr = μkFN = (.7)(1470) = 1029 N = 1000 N

1000 N

μs = .62, μk = .48What is the acceleration if there is a force of 72 N in the direction an 8.5 kg block is already sliding?

FFr = μkFN, FN = mg, FFr = μkmgFFr = (.48)(8.5 kg)(9.8 N/kg) = 39.984 NF = ma<72 N - 39.984 N> = (8.5 kg)a, a = 3.77 = 3.8 ms-2

3.8 m/s/s

72 N8.5 kg

vFFr

μs = .62, μk = .48A 22 kg block is sliding on a level surface initially at 12 m/s. What time to stop?

FFr = μkFN, FN = mg, FFr = μkmgFFr = (.48)(22 kg)(9.8 N/kg) = 103.488 NF = ma< -103.488 N> = (22 kg)a, a = -4.704 ms-2

v = u + at, v = 0, u = 12, a = -4.704 ms-2, t = 2.55 s = 2.6 s

2.6 s

22 kg

v=12m/sFFr

μs = .62, μk = .48

A 6.5 kg box accelerates and moves to the right at 3.2 m/s/s, what force must be applied?

FFr = μkFN, m = 6.5 kgFFr = (.48)(6.5 kg)(9.8 N/kg) = 30.576 NF = ma< F - 30.576 N > = (6.5 kg)(3.2 ms-2), F = 51.376 = 51 N

51 N

F = ?FFr

6.5 kg

v a = 3.2 ms-2

μs = .62, μk = .48A 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 = -5.7143 ms-2

FFr = μkFN, FN = mg, FFr = μkmgFFr = (.48)(22 kg)(9.8 N/kg) = 103.488 NF = ma< -103.488 N + F> = (22 kg)(-5.7143 ms-2), F = -22 N (left)

-22 N (to the left)

22 kg

v=12m/sFFr F = ?

μs = .62, μk = .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?

FFr = μkFN, FN = mg, FFr = μkmgFFr = (.48)m(9.8 ms-2) = m(4.704 ms-2)F = ma< 35 N - m(4.704 ms-2) > = m(1.2 ms-2)35 N = m(4.704 ms-2) + m(1.2 ms-2) = m(4.704 ms-2 + 1.2 ms-2)m = (35 N)/(5.904 ms-2) = 5.928 kg = 5.9 kg

5.9 kg

35 NFFrm

v a = 1.2 ms-2