A block of mass m 1 = 2.15 kg and a block of mass m 2 = 5.85 kg are connected by

Question

A block of mass m1 = 2.15 kg and a block of mass m2 = 5.85 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. The fixed, wedge-shaped ramp makes an angle of = 30.0° as shown in the figure. The coefficient of kinetic friction is 0.360 for both blocks.

(a) Draw force diagrams of both blocks and of the pulley.

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(b) Determine the acceleration of the two blocks. (Enter the magnitude of the acceleration.)
m/s2

(c) Determine the tensions in the string on both sides of the pulley.

Explanation / Answer

m2 provides the motive force:

Fm = m*g*sin = 5.85kg * 9.8m/s² * sin30.0º = 28.665 N

and both blocks resist motion with friction:

Ff = µ*m1*g + µ*m2*g*cos

Ff = 0.360 * 9.8m/s² * (2.15 + 5.85*cos30.0º)kg = 25.46 N

for a net force Fnet = Fm - Ff = 3.206 N

The equivalent mass of the pulley is Meq = ½ * 10.0kg = 5 kg

(since its moment of inertia is I = ½mr²)

so the total mass

m = m1 + m2 + Meq = 13 kg

(b) acceleration a = Fnet / m = 3.206N / 13kg = 0.247 m/s²

(c) Left side:

m1 * a = T1 - µ*m1*g

2.15kg * 0.247m/s² = T1 - 0.360 * 2.15kg * 9.8m/s²

T1 = 8.12 N

Right side:

m2*a = m2*g*sin - T2 - µ*m2*g*cos

5.85kg * 0.247m/s² = 5.85kg * 9.8m/s² * (sin30.0º - 0.360 * cos30.0º) - T2

T2 = 9.35 N

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