A block of mass m 1 = 1 kg rests on a table with which it has a coefficient of f

Question

A block of mass m1= 1 kg rests on a table with which it has a coefficient of friction = 0.55. A string attached to the block passes over a pulley to a block of mass m3= 3 kg. The pulley is a uniform disk of mass m2= 0.5 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

1)

With what acceleration does the mass m3 fall?

a =

2)

What is the tension in the horizontal string, T1?

T 1 =

3)

What is the tension in the vertical string, T3?

T 3 =

A block of mass m1= 1 kg rests on a table with which it has a coefficient of friction ½ = 0.55. A string attached to the block passes over a pulley to a block of mass m3= 3 kg. The pulley is a uniform disk of mass m2= 0.5 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley. 1) With what acceleration does the mass m3 fall? a = 2) What is the tension in the horizontal string, T1? T 1 = 3) What is the tension in the vertical string, T3? T 3 =

Explanation / Answer

1) m3g - T3 = m3a====> T3 = m3*(g-a)...........1

T1 -uk*m1g = m1a====> T1 = m1*(ukg+a)...........2

(T3 - T1)*R = I*alfa = 0.5*m2*R^2*a/R

T3 - T1 = 0.5*m2*a..........3

m3*g - m3*a - uk*m1*g - m1*a = 0.5*m2*a

a = (m3*g - uk*m1*g)/(m3+m2+ukm1)

a = ((3*9.8)-(0.55*1*9.8))/(3+.5+(0.55*1)) = 5.92 m/s^2

2) T1 = 1*((0.55*9.8)+5.92) = 11.31 N

3) T3 = 3*(9.8-5.92) = 11.64

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