4.2.1 Translational Equilibrium

A body can remain at its current position only if all the external forces acting on it sum up to zero. For a body to be in translational equilibrium, it is necessary that

\displaystyle \sum{F}=0


When a 20 N weight is hung on a clothes line, the line sags as shown in diagram. Determine the tension in the clothes line.


Method 1

Consider the equilibrium of the section of the rope where the mass is hung.

Horizontally, SFx = 0:               \displaystyle \begin{aligned}{{T}_{1}}\sin 75{}^\circ &={{T}_{2}}\sin 55{}^\circ \\{{T}_{1}}&=0.8480{{T}_{2}}\end{aligned}

Vertically, SFy = 0:                   \displaystyle \begin{aligned}{{T}_{1}}\cos 75{}^\circ +{{T}_{2}}\cos 55{}^\circ &=20\\(0.8480{{T}_{2}})\cos 75{}^\circ +{{T}_{2}}\cos 55{}^\circ &=20\\{{T}_{2}}&=25.2\text{ N}\end{aligned}

Substitute back into earlier equation: \displaystyle {{T}_{1}}=0.8480(25.2)=21.4\text{ N}

Method 2

Because the three forces should add up to zero, they should form a closed vector triangle.

Using sine rule:          

\displaystyle \frac{W}{{\sin 50{}^\circ }}=\frac{{{{T}_{1}}}}{{\sin 55{}^\circ }}=\frac{{{{T}_{2}}}}{{\sin 75{}^\circ }}


50 g lifts 100 g

Example Problems

Picture Frame

Block on Slop

Buoy in River

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