Question

Remarks It's interesting that the final result depends only on the coefficient of static friction. Notice also how similar Equations (1) and (2) are to the equations developed in previous problems. Recognizing such patterns is key to solving problems successfully.

Question A larger static friction constant would result in a: (Select all that apply.)

a) smaller component of gravitational force along the ramp at the maximum angle.

b) larger maximum angle.

c)smaller maximum angle.

d)larger component of gravitational force along the ramp at the maximum angle.

e)larger component of normal force at the maximum angle.

PRACTICE IT Use the worked example above to help you solve this problem. Suppose a block with mass 2.80 kg is resting on a ramp. If the coefficient of static friction between the block and the ramp is 0.420, what maximum angle can the ramp make with the horizontal before the block starts to slip down?

EXERCISE Use the values from PRACTICE IT to help you work this exercise. The ramp in the figure is roughed up and the experiment repeated.

(a) What is the new coefficient of static friction if the maximum angle turns out to be 38.0°?

(b) Find the maximum static friction force that acts on the block.
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Answer 1

Question: A larger static friction constant would result in a: (Select all that apply.)

Answer: (b) and (d) options (Reason: With increase in the angle, Sin(θ) increases, while Cos(θ) decreases. Therefore gravitation force along the ramp increases because it is directly proportional to Sin(θ). And the force normal to the ramp decreases)

En = Fg sino PRACTICE IT Gravitational Force Fg = mxg = (2.8 x 9.81) = 27.5 N For max. angle without sliding: Fn = Ft (i.e. Horizontal component of gravity = Frictional force) Fg Sine = Fnxu 27.5 x Sin = 27.5.Cosexu Sino Cose Tane = 0.42 0 = Tan -0.42) = 22.8° EXERCISE (a) New max. angle = 38° New coefficient of friction u = Tan(38) = 0.78 (b) Max. static friction = Fi = Fnxu = Fgx Cose x u = 27.5 x Cos(38°) x 0.78 = 16.9 N

En = Fg sino PRACTICE IT Gravitational Force Fg = mxg = (2.8 x 9.81) = 27.5 N For max. angle without sliding: Fn = Ft (i.e. Horizontal component of gravity = Frictional force) Fg Sine = Fnxu 27.5 x Sin = 27.5.Cosexu Sino Cose Tane = 0.42 0 = Tan -0.42) = 22.8° EXERCISE (a) New max. angle = 38° New coefficient of friction u = Tan(38) = 0.78 (b) Max. static friction = Fi = Fnxu = Fgx Cose x u = 27.5 x Cos(38°) x 0.78 = 16.9 N