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Chapter 7: Q78P (page 149)

In Fig. 5-64, a force of magnitude 12N is applied to a FedEx box of mass m2=1.0kG . The force is directed up a plane tilted by 胃=37掳 . The box is connected by a cord to a UPS box of mass m1=3.0kg on the floor. The floor, plane, and pulley are frictionless, and the masses of the pulley and cord are negligible. What is the tension in the cord?

Short Answer

Expert verified

Tension (T) is 4.6N

Step by step solution

01

Given information

It is given that,

F=12N

m2=1.0kg

胃=370

m1=3.0kg

02

Determining the concept

The problem is based on Newton鈥檚 second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagramdifferent forces acting on the objects can be found. Further, using Newton's second law of motion, the acceleration and tension in the cord can be computed.

Formula:

According to the Newton鈥檚 second law of motion,

Fnet=鈭�Ma

03

Determining the tension in the cord

Free Body Diagram:

F-I-m2gsin胃=m2a

T=m1a

By solving above two equations,

a=F-m2gsin胃m1+m2

a=12-(1x9.8xsin(37掳)(3+1)=1.53m/s2

Now, tension is given by,

T=m1a=3x1.53=4.6N

Hence, tension is 4.6N

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