- #1

MissBisson

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## Homework Statement

A 27.1

*kg*block (m1) is on a horizontal surface, connected to a 6.70

*kg*block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.097

*m*and a moment of inertia I=0.070

*kgm2*. A force F = 213.7

*N*acts on m1 at an angle theta = 34.5°. There is no friction between m1 and the surface. What is the upward acceleration of m2?

## Homework Equations

Torque = F x R

Torque = I x alpha

a = R x alpha

I = 1/2 m x R^2

## The Attempt at a Solution

Block 1 : Fcos(theta) - T1 = m1a T1 = Fcos(theta) - m1a (opposite direction of accereration)

Block 2 : T2 - m2g = m2a T2 = m2a + m2g (in the direction of accleration)

Torque = (T2-T1)R = I x (a/R)

(m2a + m2g - Fcos(theta) + m1a)R = 1/2 m x R^2 x (a/R)

m2a + m2g - Fcos(theta) + m1a = (m x R^2 x a)/(2 x R^2) R^2 cancel out

m2a + m2g - Fcos(theta) + m1a = (m x a)/2

m2a + m1a - (m x a)/2 = Fcos(theta) - m2g

a (m2 + m1 - m/2) = Fcos(theta) - m2g

a = (Fcos(theta) - m2g)/(m2 + m1 - m/2) -----------------> A

a = (213.7cos(34.5) - 6.7*9.81)/ (6.7 + 27.1 - 17.88/2)

a = (110.39/26.36)

a= 4.187m/s^2

I know that the answer is 2.68m/s^2 and that the equation A should be

a = (Fcos(theta) - m2g)/(m2 + m1 + m/2)

But i don't know how to get that...