In earlier days, horses pulled barges down canals in the manner shown in Figure. Suppose the horse pulls on the rope with a force of$\mathbf{7900}\mathbf{}\mathbf{N}$at an angle of $\mathbf{\theta }\mathbf{=}\mathbf{}\mathbf{18}\mathbf{°}\mathbf{}$to the direction of motion of the barge, which is headed straight along the positive direction of an xaxis. The mass of the barge is$\mathbf{}\mathbf{9500}\mathbf{}\mathbf{kg}$, and the magnitude of its acceleration is $\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. (a) What is the magnitude and (b) What is the direction (relative to positive x) of the force on the barge from the water?

• The tension in the rope is,$\mathrm{T}=7900\mathrm{N}.$

• The angle made by the rope with the direction of motion of the barge is,$\mathrm{\theta }={18}^0$ .

The force on the barge is equal to the product of its mass and acceleration. If the force is not in the direction of the acceleration, we can resolve the force to find its component along with the acceleration. The component of the force in the direction of acceleration is responsible for the acceleration of the object. The free-body diagram can help us to understand the direction of forces, and their components acting on the body.

Formulae:

The force F acting on the body is given as,

$\sum \mathrm{F}=\mathrm{ma}$

Where, m is mass of the body, a is the acceleration of the body.

Consider unknown force F whose components are${\mathrm{F}}_{\mathrm{x}}\hspace{0.33em}\mathrm{and}\hspace{0.33em}{\mathrm{F}}_{\mathrm{y}}.$

Now apply Newton’s law along x and y directions considering the net force equal to mass times acceleration in x direction and zero in y direction.

The force equation along the x-axis can be written as,

$$\begin{gathered} \mathrm{Tcos}\left({18}^{\circ }\right)+{\mathrm{F}}_{\mathrm{x}}=\mathrm{ma} \\ 7900\mathrm{N}.\cos \left({18}^{\circ }\right)+{\mathrm{F}}_{\mathrm{x}}=9500\mathrm{kg}\times 0.12\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{F}}_{\mathrm{x}}=-6.4\times {10}^3\mathrm{N} \end{gathered}$$

Hence, the force in the x-direction is, ${\mathrm{F}}_{\mathrm{x}}=-6.4\times {10}^3\mathrm{N}$

Along the y-axis, the force equation can be written as,

role="math" localid="1657167114006" $$\begin{gathered} \mathrm{Tsin}\left({18}^{\circ }\right)+{\mathrm{F}}_{\mathrm{y}}=0 \\ 7900\sin \left({18}^{\circ }\right)+{\mathrm{F}}_{\mathrm{y}}=0 \\ {\mathrm{F}}_{\mathrm{y}}=2.4\times {10}^3\mathrm{N} \end{gathered}$$

Hence, the force in the y-direction is, ${\mathrm{F}}_{\mathrm{y}}=-2.4\times {10}^3\mathrm{N}$

The magnitude of force can be found by using the formula,

$\mathrm{F}=\sqrt{{\mathrm{F}}_{\mathrm{x}}^2+{\mathrm{F}}_{\mathrm{y}}^2}$

Substitute the values of the component of forces in the above equation, we get

$$\begin{gathered} \mathrm{F}=\sqrt{{\left(-6.4\times {10}^3\right)}^2+{\left(-2.4\times {10}^3\right)}^2} \\ =6.8\times {10}^3\mathrm{N} \end{gathered}$$

Therefore, the magnitude of force is $6.8\times {10}^3\hspace{0.33em}\mathrm{N}$.

Direction for force,

$$\begin{gathered} \mathrm{\varphi }={\tan }^{-1}\left(\frac{{\mathrm{F}}_{\mathrm{y}}}{{\mathrm{F}}_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{-2.4\times {10}^3\mathrm{N}}{-6.4\times {10}^3\mathrm{N}}\right) \\ ={21}^{°} \end{gathered}$$

Both the components of foce are in third quadrant hence, direction would be $(180°+21°)=201°\mathrm{from}+\mathrm{x}-\mathrm{axis}.$