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Chapter 12: Problem 14

Conditions for Equilibrium A body is subject to three forces: \(\vec{F}_{1}=2 \hat{\imath}+2 \hat{\jmath} \mathrm{N},\) applied at the point \(x=2 \mathrm{m}, y=0 \mathrm{m} ; \vec{F}_{2}=-2 \hat{\imath}-3 \hat{\jmath} \mathrm{N},\) applied at \(x=-1 \mathrm{m}, y=0 \mathrm{m} ;\) and \(\vec{F}_{3}=1 \hat{j} \mathrm{N},\) applied at \(x=-7 \mathrm{m}\) \(y=1 \mathrm{m} .\) Show that (a) the net force and (b) the net torque about the origin are both zero.

Short Answer

Expert verified
The net force on the body is 0 N and the net torque on the body about the origin is 0 Nm. Hence, both conditions for equilibrium are satisfied, thus the body is in equilibrium.

Step by step solution

01

Sum of Force Vectors

Firstly, calculate the net force on the body by adding up the force vectors. This is done by adding the corresponding components of the force vectors: \( \vec{F}_{net} = \vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3} = (2 \hat{\imath}+2 \hat{\jmath})+(-2 \hat{\imath}-3 \hat{\jmath})+(0 \hat{\imath}+1 \hat{\jmath}) = 0 \hat{\imath}+0 \hat{\jmath} \) N.
02

Calculation of Torque

Secondly, calculate the torque caused by each force about the origin. The torque can be calculated by using the formula \( \vec{τ} = \vec{r} \times \vec{F} \). Here, \( \vec{r} \) is the position vector of the point where the force is applied. Torques: \( \vec{τ}_{1} = (2 \hat{\imath}) × (2 \hat{\imath}+2 \hat{\jmath}) = 0 N m, \vec{τ}_{2} = (-1 \hat{\imath}) × (-2 \hat{\imath}-3 \hat{\jmath}) = 3 N m, \vec{τ}_{3} = (-7 \hat{\imath}+1 \hat{\jmath}) × (0 \hat{\imath}+1 \hat{\jmath}) = -7 N m.
03

Sum of Torques

Add up the torques to get the net torque on the body. Net torque is \( \vec{τ}_{net} = \vec{τ}_{1}+\vec{τ}_{2}+\vec{τ}_{3} = 0 Nm+3 Nm-7 Nm = 0 Nm\).

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