Chapter 6, Problem 8PS

(a)

To determine

To compare: the work done by ${\text{F}}_1$ with the work done by ${\text{F}}_2$ in moving along the vector $\overline{PQ}$ .

Answer to Problem 8PS

Both work done are same

Explanation of Solution

Given information:

Two forces of the same magnitude ${\text{F}}_1$ and ${\text{F}}_2$ act an angles ${\theta }_1$ and ${\theta }_2$ respectively,

When ${\theta }_1=-{\theta }_2$

Calculation:

The work done by a force through displacement $\text{d}$ is given below.

  $\text{W=F}\text{.d=Fdcos}\theta$

The work for both forces through a displacement $PQ$ . Since the cosine for both the cases is same and the magnitudes of the forces and the displacements are same, the work done by the forces is same.

(b)

To determine

To compare: the work done by ${\text{F}}_1$ with the work done by ${\text{F}}_2$ in moving along the vector $\overline{PQ}$ .

Answer to Problem 8PS

The work done by ${\text{F}}_2$ is $\sqrt{3}$ times that of ${\text{F}}_1$ .

Explanation of Solution

Given information:

Where

  ${\theta }_1=60°$ and ${\theta }_2=30°$ .

Calculation:

The work for both forces through a displacement $PQ$ has been given in the form of the figure below. Since the magnitude of the forces and the displacements are same, the work done by ${\text{F}}_2$ is $\sqrt{3}$ times that of ${\text{F}}_1$ .