Question

The force $\overrightarrow{\mathbf{F}}$ in Fig. 12-70 keeps the $\mathbf{6}\mathbf{.40}\mathbf{\text{\hspace{0.17em}kg}}$ block and the pulleys in equilibrium. The pulleys have negligible mass and friction. Calculate the tension Tin the upper cable. (Hint:When a cable wraps halfway around a pulley as here, the magnitude of its net force on the pulley is twice the tension in the cable.)

Short answer

Tension T in the upper cable is $71.68\text{\hspace{0.17em}N}$.

Step-by-step solution

Understanding the given information

Mass of the block, $\text{m}=6.40\text{\hspace{0.17em}kg}$

Concept and formula used in the given question

Using the equilibrium conditions, you can write the equations for force for the lowest pulley and find the tension in terms of applied force. You can do it for the remaining two pulleys and find the tension in the string holding the topmost pulley. The formula used is given below.

${\sum }_{}^{}{\text{F}}_{\text{net}}=0$

Calculation for the tension T in the upper cable

From the figure, we can conclude that

$\begin{array}{l}{T}_1=F\\ T_2=F+T_1\\ T_2=2F\\ T_3=2F+T_2\\ T_3=4F\\ T=4F+T_3\\ T=8F\end{array}$

In addition, we know the value of force,$\text{F}=62.72\text{\hspace{0.17em}N}$

We can write,

$\begin{array}{l}{T}_1+T_2+T_3=mg\\ 7F=mg\\ F=\frac{mg}{7}\\ F=\frac{62.72N}{7}\\ F=8.96\text{\hspace{0.17em}N}\end{array}$

We already found that,

$T=8F$

So,

$\begin{array}{c}T=\left(8.96N\right)\left(8\right)\\ =71.68\text{\hspace{0.17em}N}\end{array}$

Thus, the tension in the upper cable is $71.68\text{\hspace{0.17em}N}$.