A block of mass m 1 = 1.60 kg and a block of mass m 2 = 5.95 kg are connected by

Question

A block of mass m1 = 1.60 kg and a block of mass m2 = 5.95 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and massM = 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) Determine the acceleration of the two blocks. (Enter the magnitude of the acceleration.)

___________________m/s2

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

left of the pulley___________N

right of the pulley__________N

left of the pulley___________N

right of the pulley__________N

Explanation / Answer

Let the acceleration = a
Let the tension in string attached to block m1 = T1
Let the tension in string attached to block m2 = T2

By Newton's second Law for block m1 -
T1 - u*m1*g = m1*a
T1 = m1*a + u*m1*g

By Newton's second Law for block m2 -
m2*sin()*g - T2 - u*m2cos()*g = m2*a
T2 = m2*sin()*g -  u*m2cos()*g - m2*a

For Disk -
Net Torque = I*
T2*R - T1*R = I*a/R
a = (T2-T1) * R^2 / (M*R^2/2)
a = ( m2*sin()*g -  u*m2cos()*g - m2*a - m1*a - u*m1*g) * 2 / M
a = 2*(5.95 * sin(30)*9.8 - 0.360*5.95*cos(30)*9.8 - 5.95 a - 1.6 a - 0.360*1.6*9.8 ) / 10
a = 0.425 m/s^2
Acceleration of the two blocks. a = 0.425 m/s^2

(b)
Tension in the left of the pulley =
T1 = m1*a + u*m1*g
T1 = 1.6*(0.425 + 0.36*9.8)
T1 = 6.32 N

Tension in the Right of the pulley =
T2 = m2*sin()*g -  u*m2cos()*g - m2*a
T2 = 5.95 * sin(30)*9.8 - 0.360*5.95*cos(30)*9.8 - 5.95 *0.425
T2 = 8.45 N

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