Question

A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are µs = 0.47 and µk = 0.4, respectively. grant (dg38258) – HW#4 Ch5+6 Frictional Gravitational Forces – storr – (37573) 2 The acceleration of gravity is 9.8 m/s 2 . 32 kg µ 21◦ What is the frictional force acting on the 32 kg mass? Answer in units of N. 018 (part 2 of 3) 10.0 points What is the largest angle which the incline can have so that the mass does not slide down the incline? Answer in units of ◦ . 019 (part 3 of 3) 10.0 points What is the acceleration of the block down the incline if the angle of the incline is 29 ◦ ? Answer in units of m/s 2 .

Answer #1

1) Force F = μR (where R is normal reaction)

R = mgCosθ

= 0.47 x 32 x 9.81 x Cos 25

Frictional Force = 137.65 N

2) Limiting angle is when component of weight along plane =
Frictional force

mg Sin θ = μmgCosθ

μ = Sin θ / Cosθ = Tan θ = 0.51

Angle θ = Tan^-1 (0.47)

Largest Angle = 25.17°

3) At 29 °, force down plane = mg Sin θ and frictional force is
μ(k)mgCosθ

Net force = mg ( Sin θ - μ(k)Cosθ )

= 32 x 9.81 (Sin 29 - 0.40Cos 29)

= 313.92 (0.484 - 0.193)

= 91.06 N

Acceleration = Force / Mass = 91.06 /32 = 2.845 m /s^2

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