Three blocks are connected on the table as shown in problem 12. The coefficient

Question

Three blocks are connected on the table as shown in problem 12. The coefficient of kinetic friction between the block of mass m2and the table is 0.485. The objects have masses of m1 = 3.75 kg, m2 = 1.30 kg, and m3 = 1.90 kg, and the pulleys are frictionless. (a) Draw free-body diagrams for each of the objects. (b) Determine the acceleration of each object, including its direction. (c) Determine the tensions in the two cords. (d) If the tabletop were smooth, would the tensions increase, decrease, or remain the same? Explain.

Explanation / Answer

As vector down on m3 1.80kg*9.8m/s^2 and also down from m1 of 5*9.8. since m1 has more mass movement is going to be in that direction therefore the friction force will be opposite or to the right. friction force is normal force X coefficient of friction or 1.25kg*9.8m/s^2 now we use sum of the forces = ma force of m1 - force of m3 - force of friction on m2 = ma since the objects are attached, they will accelerate at the same rate so a is a the same for all 3 masses. and we use m1+m2+m3 as m in this equation and solve for a force of m3 + force of m1 + force of friction on m2(to the left) - force of friction on m2(to the right). this will solve the right side. you must calculate force of friction as it cannot be greater than the force applied to it left side would just be the opposite on the friction portion.

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