Two boxes one with mass m1= 4 kg and the other with mass m2= 6 kg, sit on a hori

Question

Two boxes one with mass m1= 4 kg and the other with mass m2= 6 kg, sit on a horizontal surface, connected by a massless rope. A person pulls horizontally on the 6 kg box by a force that gives the two boxes an acceleration of 0.3 m/s^2. The coefficient of kinetic friction between m1 and the surface equals 0.25 and that between the m2 and surface equals 0.15. A. Draw a free body diagram for each box. B. Find the friction force for each box. C. Find the tension on the rope. D. Find the applied force F.

Explanation / Answer

(A) Draw the free body diagram by yourself. It is very easy. Follow the following procedure -

T = Tension in the rope connecting mass m1 and m2.

F = Applied force.

mu1 = coefficient of kinetic friction between m1 and the surface.

mu2 = coefficient of kinetic friction between m2 and the surface.

Now take the mass m2 -

Applied force F will act in the right direction(--------------------->).

Frictional force mu2*m2*g in the left direction(<-----------)

Tension T in the left direction (<-----------------------)

Acceleration of the block in the right direction (-------------------->).

Repeat the same for the box m1, except T in the right direction.

(B) Frictional force on m1 = mu1*m1*g = 0.25*4*9.81 = 9.81 N

Frictional force on m2 = mu2*m2*g = 0.15*6*9.81 = 8.83 N

(C) Total frictional force of the two boxes = 9.81 + 8.83 = 18.64 N

For box m1 -

T - mu1*m1*g = m1*a

=> T - 9.81 = 4*0.3 = 1.2

=> T = 1.2+9.81 = 11.01 N

(D) Take the case of box m2 -

F - T - mu2*m2*g = m2*a

=> F - 11.01 - 8.83 = 6*0.3 = 1.8

=> F = 1.8 + 11.01 + 8.83 = 21.64 N.

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