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202202070828 Dynamics Figures (Elementary) Q1

As shown in the sketch above, a mass m is suspended from a string of length L. Gravity \mathbf{g} is acting downward in the diagram. A second string exerts a horizontal force \mathbf{F} such that m is in equilibrium at a horizontal distance x.

(a) Draw a diagram showing all forces acting on m.

(b) What is the tension in the suspension string? Give both direction and magnitude.

Extracted from R. L. Gray. (1973). Physics Problems: Mechanics and Heat.

Solution.

(a) The free body diagram is shown below:

(b) Obviously,

\begin{aligned} \mathbf{T} & =F\,\hat{\mathbf{i}}+mg\,\hat{\mathbf{j}} \\ T & =|\mathbf{T}|=\sqrt{(F)^2+(mg)^2} \\ \end{aligned}.

Or, in an old-school way, we have horizontally

F=T\cos\theta

and vertically

mg=T\sin\theta,

squaring and summing up the equations,

\begin{aligned} (F)^2+(mg)^2 & =(T\cos\theta )^2+(T\sin\theta )^2 \\ F^2+m^2g^2 & = T^2(\cos^2\theta +\sin^2\theta ) \\ F^2+m^2g^2 & = T^2 \\ T & = \sqrt{F^2+m^2g^2} \\ \end{aligned}

Dividing the equation of the vertical by the horizontal,

\begin{aligned} \frac{mg}{F} & = \frac{T\sin\theta}{T\cos\theta} \\ \frac{mg}{F} & = \tan\theta \\ \theta & = \tan^{-1}\bigg(\frac{mg}{F}\bigg) \\ \end{aligned}

the direction of tension makes an angle \theta =\tan^{-1}(\frac{mg}{F}) with the level.