A block of mass m 1 = 2.40 kg and a block of mass m 2 = 6.25 kg are connected by

Question

A block of mass m1 = 2.40 kg and a block of mass m2= 6.25 kg are connected by a masslessstring over a pulley in the shape of a solid disk having radiusR = 0.250 m and mass M = 10.0 kg. These blocksare allowed to move on a fixed block-wedge of angle = 30.0°. The coefficient of kinetic frictionis 0.360 for both blocks. Draw free-body diagrams of both blocksand of the pulley. Figure(http://www.webassign.net/pse/pse6_p10-37.gif) (a) Determine the acceleration of the twoblocks.
1 m/s2

(b) Determine the tensions in the string on both sides of thepulley. left of the pulley 2 N right of the pulley 3 N
(a) Determine the acceleration of the twoblocks.
1 m/s2

(b) Determine the tensions in the string on both sides of thepulley. left of the pulley 2 N right of the pulley 3 N left of the pulley 2 N right of the pulley 3 N

Explanation / Answer

Use newton's second law, F=ma, to write force equations foreach block and pulley. For block 1: m1a = T1 -m1g         whereT1 is the tension to the left of the pulley and isthe coefficient of friction For block 2: m2a = m2g*sin() - T2 -m2g*cos() For the pulley: I = T2R - T1R The moment of inertia for a disk: I = 1/2*MR2 Angular acceleration is equal to: = a / R
1/2*M*a = T2 - T1
Adding the three equations together, we solve foracceleration: a*(m1+m2+1/2*M) =m2g*sin() - m2g*cos() -m1g a = 9.81m/s2*(6.25kg*sin(30) - 0.36*6.25kg*cos(30)- 0.36*2.4kg) / (6.25kg + 2.4kg + 1/2*10kg)
=_____ m/s2 B) To the left of the pulley: T1 = m1*(a + g)
To the right of the pulley: T2 =  m2g*sin()- m2a - m2g*cos()

solve by plugging values
B) To the left of the pulley: T1 = m1*(a + g)
To the right of the pulley: T2 =  m2g*sin()- m2a - m2g*cos()

solve by plugging values

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