Chapter 4, Problem 52P

(a)

To determine

The unknown tension and masses.

Answer to Problem 52P

The value of the tension $T_1$ is $60\text{N}$ , $T_2$ is $52\text{N}$ and the value of the mass is $m$ is $5.3\text{kg}$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Formula used:

For the given diagram, the expression for the horizontal forces is given by,

  $\begin{array}{l}{T}_1\cos 60°=T_3\\ T_1=\frac{T_3}{\cos 60°}\end{array}$

The equation for the forces resolved in the vertical direction is given by,

  $T_2=T_1\sin 60°$

The expression for the mass of the block is given by,

  $m=\frac{T_2}{g}$

Calculation:

The tension $T_1$ is calculated as,

  $\begin{array}{c}{T}_1=\frac{T_3}{\cos 60°}\\ =\frac{60\text{N}}{\cos 60°}\\ =60\text{N}\end{array}$

The expression for the value of the tension $T_2$ in the rope is calculated as,

  $\begin{array}{c}{T}_2=T_1\sin 60°\\ =\left(60\text{N}\right)\sin 60°\\ =52\text{N}\end{array}$

The mass of the block is calculated as,

  $\begin{array}{c}m=\frac{T_2}{g}\\ =\frac{52\text{N}}{9.8\text{m}/{\text{s}}^2}\\ =5.3\text{kg}\end{array}$

Conclusion:

Therefore, the value of the tension $T_1$ is $60\text{N}$ , $T_2$ is $52\text{N}$ and the value of the mass is $m$ is $5.3\text{kg}$ .

(b)

To determine

The unknown tension and masses.

Answer to Problem 52P

The value of the tension $T_1$ is $42.6\text{N}$ , $T_2$ is $46.2\text{N}$ and the value of the mass is $m$ is $4.7\text{kg}$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (b) is shown in Figure 2

  

Figure 2

For the given diagram, the expression for the horizontal forces is given by,

  $\begin{array}{l}{T}_1\sin 60°=T_3\sin 60°\\ T_1=\frac{T_3}{\tan 60°}\end{array}$

The equation for the forces resolved in the vertical direction is given by,

  $T_2=T_3\sin 60°-T_1\cos 60°$

The expression for the mass of the block is given by,

  $m=\frac{T_2}{g}$

Calculation:

The tension $T_1$ is calculated as,

  $\begin{array}{c}{T}_1=\frac{T_3}{\tan 60°}\\ =\frac{80\text{N}}{\tan 60°}\\ =42.6\text{N}\end{array}$

The expression for the value of the tension $T_2$ in the rope is calculated as,

  $\begin{array}{c}{T}_2=T_3\sin 60°-T_1\cos 60°\\ =\left(80\text{N}\right)\sin 60°-\left(42.6\text{N}\right)\cos 60°\\ =46.2\text{N}\end{array}$

The mass of the block is calculated as,

  $\begin{array}{c}m=\frac{T_2}{g}\\ =\frac{46.2\text{N}}{9.8\text{m}/{\text{s}}^2}\\ =4.7\text{kg}\end{array}$

Conclusion:

Therefore, the value of the tension $T_1$ is $42.6\text{N}$ , $T_2$ is $46.2\text{N}$ and the value of the mass is $m$ is $4.7\text{kg}$ .

(c)

To determine

The unknown tension and masses.

Answer to Problem 52P

The value of the tension $T_1$ is $34\text{N}$ , $T_2$ is $58.8\text{N}$ and the value of the mass is $m$ is $3.47\text{kg}$ .

Explanation of Solution

Formula used:

The free body diagram for the tension in the rope for figure (c) is shown in Figure 3

  

Figure 3

For the given diagram, the expression for the horizontal forces is given by,

  $T_2=m_{1}g$

The equation for the forces resolved in the vertical direction is given by,

  $T_2=T_1\sin 60°+T_3\sin 60°$

The expression for the mass of the block is given by,

  $m=\frac{T_1}{g}$

The expression for the forces resolved in the horizontal direction is given by,

  $\begin{array}{l}{T}_1\cos 60°=T_3\cos 60°\\ T_1=T_3\end{array}$

Calculation:

The tension $T_2$ is calculated as,

  $\begin{array}{c}{T}_2=m_{1}g\\ =\left(6\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =58.8\text{N}\end{array}$

The expression for the value of the tension $T_1$ in the rope is calculated as,

  $\begin{array}{l}{T}_2=T_1\sin 60°+T_3\sin 60°\\ 58.8\text{N}=T_1\sin 60°+T_1\sin \sin 60°\\ T_1=\frac{58.8\text{N}}{2\sin 60°}\\ T_1=34\text{N}\end{array}$

The mass of the block is calculated as,

  $\begin{array}{c}m=\frac{T_1}{g}\\ =\frac{34\text{N}}{9.8\text{m}/{\text{s}}^2}\\ =3.47\text{kg}\end{array}$

Conclusion:

Therefore, the value of the tension $T_1$ is $34\text{N}$ , $T_2$ is $58.8\text{N}$ and the value of the mass is $m$ is $3.47\text{kg}$ .