What is the tension in the rope that connects the boxes? What is the mass m of the second box?

Two boxes connected by a light horizontal rope are on a horizontal surface. The coefficient of kinetic friction between each box and the surface is .3. One box (box B) has a mass 4.65 kg, and the other box (box A) has mass *m*. A force F with magnitude 40.3 N and direction 53.1 degrees above the horizontal is applied to the 4.65 kg box, and both boxes move to the right with 1.5 m/s^2.

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When the boxes are dragged, they move with the same speed and acceleration.

The combined drag on the boxes is 0.3(m + 4.65)*9.8 where 0.3 is the coefficient of kinetic friction and 9.8 is the acceleration due to gravity. (m+4.65)*9.8 is the total downward force, and the force due to friction is 0.3 of this.

To work out the horizontal forward force applied to box B F1 on box B use cos(53.1) = F1/40.3.

So F1 = 40.3 x cos(53.1) = 24.2N

The resultant forward force on both boxes is (using force = mass x acceleration)

(m + 4.65)*1.5

This is also equal to the forward force applied to the boxes minus the force due to friction.

So (m + 4.65)*1.5 = 24.2 - 0.3*(m + 4.65)*9.8

Gathering terms this gives

(9.8(0.3) + 1.5)m = 24.2 - 0.3(4.65)9.8 - 4.65(1.5)

m = 3.554/4.44 = 0.800kg

To work out the tension on the rope, we consider the forces acting on box A only.

The resultant forward force is mass x acceleration = 1.5mN = 1.5(0.8)N = 1.2N

The force due to friction or drag is 0.3(9.8)m = 0.3(9.8)0.8 = 2.352N

So the tension T satisfies

T - 2.352N = 1.2N

`therefore` T = 1.152N

**T****ension = 1.15N and m = 0.800kg**

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