Question

Use the second condition for equilibrium(net $\mathit{\tau }\mathbf{=}\mathbf{0}$)to calculate${\mathit{F}}_{\mathbf{P}}$inExample 9.1, employing any data given or solved for in part (a) of the example.

Short answer

The calculated force is $F_P=568\mathrm{N}$.

Step-by-step solution

Given data

• The mass is,$m_1=26\mathrm{kg}$.
• The distance from the hinge is,$r_1=1.60m$.
• The mass of the other is, $m_2=32\mathrm{kg}$.

• The distance of the other from the hinge is,$r_2=1.30\mathrm{m}$.

Introduction of force

When the system is in equilibrium, the net force acting on a body will be equal to zero. In other words, the forces will balance each other, and the system will remain in equilibrium as long as no external disturbance is there.

Calculation of the force

The net torque is zero; we can write the equation as,

$$\begin{gathered} \underset{}{\sum \tau }=0 \\ {F}_P\times r_1-W_2\left(r_1+r_2\right)=0 \\ {F}_p\times r_1-m_{2}g\left(r_1+r_2\right)=0 \end{gathered}$$

Here g is the gravitational acceleration.

Substitute the values in the above expression, and we get,

$$\begin{gathered} F_P\times 1.60=32\times 9.8\times \left(1.60+1.30\right) \\ {F}_P=\frac{32\times 9.8\times \left(1.60+1.30\right)}{1.60} \\ {F}_P=568\mathrm{N} \end{gathered}$$

Hence, the force is$F_P=568\mathrm{N}$.