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Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate A has mass mA, and crate B has mass mB. The coefficient of kinetic friction between each crate and the surface is μk. The crates are pulled to the right at constant velocity by a horizontal force F→. Draw one or more free-body diagrams to calculate the following in terms of mA, mB and μk: (a) the magnitude of F→ and (b) the tension in the rope connecting the blocks.

Figure E5.37

Short Answer

Expert verified
  1. The magnitude of F→is, μkgmB+mA.
  2. The tension in the connecting rope is, μkmAg.

Step by step solution

01

Identification of given data

The given data can be listed below as,

The mass of crate A is, mA.

The mass of crate B is, mB.

The horizontal force is, F→.

The coefficient of kinetic friction is, μk.

02

Significance of tension

Tension is a pulling force which transmitted axially with the help of rope, string or cable at the end is known as tension.

03

(a) Determination of magnitude of F→.


The expression for frictional force on crate A and b can be expressed as,

fA=μkmAgfB=μkmBg

Here μk is the coefficient of kinetic friction, mA is the mass of the crate A, mB is the mass of the crate B.

The expression for force using Newton’s second law can be expressed as,

F→-fB-fA=0

Here fA is the force on crate A, fB is the force on crate B.

Substitute μkmAg for fA, μkmBgfor fB in above equation.

F→-μkmAg-μkmBg=0F→=μkgmA+mB

Hence, the required magnitude of F→is, μkgmA+mB.

04

(b) Determination of tension in the rope connecting the blocks.

For Tension considering the block rare one block A and block B as an isolated system, the expression for tension can be expressed as,

T-μkmAg=0T=μkmAg

Hence, the required tension is, μkmAg.

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