Three blocks are connected on the table as shown below. The coefficient of kinet

Question

Three blocks are connected on the table as shown below. The coefficient of kinetic friction between the block of mass m_2 and the table is 0.325. The objects have masses of m_1 = 3.75 kg, m_2 = 1.20 kg, and m_3 = 2.10 kg, and the pulleys are frictionless. Draw free-body diagrams for each of the objects. Determine the acceleration of each object, including its direction. m_1 magnitude___m/s^2 m_1 direction m_2 magnitude m_2 direction m_3 magnitude m_3 direction Determine the tensions in the two cords. left cords right cord If the tabletop were smooth, would the tensions increase, decrease, or remain the same? The tension in the left cord will decrease while the tension in the right cord will increase. The tension in the left cord will increase while the tension in the right cord will decrease. The tensions in the both cords will remain the same.

Explanation / Answer

A for m1 is down for m2 is to the left and m3 is up

Free body diagrams about each yields

m1*g - Tleft = m1*a so T (left) = m1*g - m1*a

T (left) - T (right) - *m2*g = m2*a

T (right) - m3*g = m3*a So T (right) = m3*g + m3*a

Now sub the first and third into the second

m1*g - m1*a - m3*a - m3*g - *m2*g = m2a

So (m1 + m2 + m3)*a = g*(m1 - m3 - *m2)

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

a = 9.8*(3.75 - 2.10 - 0.325*1.2)/(3.75 + 1.2 + 2.1)

a = 1.75 m/s^2 The directions were given above

Now T (left) = m1*g - m1*a = 3.75*(9.8 - 1.75) = 30.18N

T (right) = m3*g + m3*a = 2.10*(9.8 + 1.75) = 24.255N

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