91Ó°ÊÓ

• First force on the body,${\mathrm{F}}_1=6.00\mathrm{N}$
• Angle made by the first force, ${\mathrm{θ}}_1=30°$
• Second force on the body, ${\mathrm{F}}_2=7.00\mathrm{N}$
• Angle made by the second force, ${\mathrm{θ}}_2=30°$
• The velocity of the object is $\stackrel{¯}{\mathrm{v}}=\left(13.0\right)\stackrel{Áåœ}{\mathrm{i}}-\left(14.0\right)\stackrel{Áåœ}{\mathrm{j}}$

Free body diagrams represent the forces acting on the object. It can be used to find the direction and magnitude of the forces acting on the object. The net force on the object can also be found by using Newton’s law of motion which states net force is equal to the vector sum of all the forces.

Draw the free-body diagram of the block, which gives x and y components of the $\stackrel{\mathbf{â‡\P Ä}}{{\mathbf{F}}_{\mathbf{1}}}\mathbf{}\mathbf{and}\mathbf{}\stackrel{\mathbf{â‡\P Ä}}{{\mathbf{F}}_{\mathbf{2}}}$. Using Newton’s second law of motion, find out$\stackrel{\mathbf{â‡\P Ä}}{{\mathbf{F}}_{\mathbf{3}}}$acting on the lemon half when it is stationary. When it is moving with constant velocity, it has zero acceleration. Hence$\stackrel{\mathbf{â‡\P Ä}}{{\mathbf{F}}_{\mathbf{3}}}$remains the same as the previous one. Due to varying velocities, the lemon half has acceleration. We can find acceleration by differentiating$\stackrel{\mathbf{→}}{\mathbf{v}}$ with respect to the time.

Formulae:

$$\begin{gathered} {\stackrel{\mathbf{â‡\P Ä}}{\mathbf{F}}}_{\mathbf{net}}\mathbf{}\mathbf{=}\mathbf{m}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{a}} \\ \stackrel{\mathbf{â‡\P Ä}}{\mathbf{a}}\mathbf{=}\frac{\mathbf{d}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}{\mathbf{dt}} \end{gathered}$$

Here, ${\stackrel{\mathbf{â‡\P Ä}}{\mathbf{F}}}_{\mathbf{net}}$is the net force, m is the mass, a is acceleration, and $\stackrel{\mathbf{¯}}{\mathbf{v}}$ is the velocity of the body.

According to the free body diagram, the x and y components of${\mathrm{F}}_1\mathrm{are}{\mathrm{F}}_{\mathrm{tx}}=-{\mathrm{F}}_1{\mathrm{³¦´Ç²õθ}}_1\mathrm{and}{\mathrm{F}}_{1\mathrm{y}}={\mathrm{F}}_1{\mathrm{²õ¾\pm ²Ôθ}}_1$.

The x and y components of${\mathrm{F}}_2\mathrm{are}{\mathrm{F}}_{2\mathrm{x}}={\mathrm{F}}_2{\mathrm{²õ¾\pm ²Ôθ}}_2\mathrm{and}{\mathrm{F}}_{2\mathrm{y}}=-{\mathrm{F}}_2{\mathrm{³¦´Ç²õθ}}_2$

localid="1657091009508" $$\begin{gathered} \stackrel{â‡\P Ä}{{\mathrm{F}}_1}=-{\mathrm{F}}_1{\mathrm{³¦´Ç²õθ}}_1+{\mathrm{F}}_1{\mathrm{²õ¾\pm ²Ôθ}}_1 \\ \stackrel{â‡\P Ä}{{\mathrm{F}}_2}=-{\mathrm{F}}_2{\mathrm{²õ¾\pm ²Ôθ}}_2+{\mathrm{F}}_2{\mathrm{³¦´Ç²õθ}}_2 \end{gathered}$$

The lemon half is stationary hence it has no acceleration. According to Newton’s second law, the net force acting on the lemon half is zero.

$$\begin{gathered} {\stackrel{â‡\P Ä}{F}}_{net}=m\stackrel{¯}{a} \\ \stackrel{â‡\P Ä}{F_1}+\stackrel{â‡\P Ä}{F_2}+\stackrel{â‡\P Ä}{F_3}=0 \\ -F_1\cos {\mathrm{θ}}_1\stackrel{Áåœ}{i+F_1}\sin {\mathrm{θ}}_1\stackrel{Áåœ}{j}+F_2\sin {\mathrm{θ}}_2\stackrel{Áåœ}{i}-F_2\cos {\mathrm{θ}}_2\stackrel{Áåœ}{j}+{\stackrel{â‡\P Ä}{F}}_3=0 \end{gathered}$$

Therefore,

$\stackrel{â‡\P Ä}{{\mathrm{F}}_3}={\mathrm{F}}_1{\mathrm{³¦´Ç²õθ}}_1\stackrel{Áåœ}{\mathrm{i}}-{\mathrm{F}}_1{\mathrm{²õ¾\pm ²Ôθ}}_1\stackrel{Áåœ}{\mathrm{j}}-{\mathrm{F}}_2{\mathrm{²õ¾\pm ²Ôθ}}_2\stackrel{Áåœ}{\mathrm{i}}+{\mathrm{F}}_2{\mathrm{³¦´Ç²õθ}}_2\stackrel{Áåœ}{\mathrm{j}}$

Now, substitute the values of forces and angle in the above equation.

$$\begin{gathered} \stackrel{â‡\P Ä}{{\mathrm{F}}_3}=\left(6\cos 30\right)\stackrel{Áåœ}{\mathrm{i}}-\left(6\sin 30\right)\stackrel{Áåœ}{\mathrm{j}}-\left(7\sin 30\right)\stackrel{Áåœ}{\mathrm{i}}+\left(7\cos 30\right)\stackrel{Áåœ}{\mathrm{j}} \\ =\left(1.70\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{i}}+\left(3.06\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{j}} \end{gathered}$$

Therefore, the third force acting on the object is $\left(1.70\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{i}}+\left(3.06\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{j}}$

When lemon half has constant velocity, so it has zero acceleration. The third force for it remains the same as,

$\stackrel{â‡\P Ä}{{\mathrm{F}}_3}=\left(1.70\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{i}}+\left(3.06\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{j}}$

The acceleration due to varying force is.

$\stackrel{â‡\P Ä}{\mathrm{a}}=\frac{\mathrm{d}\stackrel{â‡\P Ä}{\mathrm{v}}}{\mathrm{dt}}$

Substitute the equation for velocity to find the acceleration.

$$\begin{gathered} \stackrel{â‡\P Ä}{\mathrm{a}}=\frac{\mathrm{d}\left(\left(13.0\mathrm{t}\right)\stackrel{Áåœ}{\mathrm{i}}-\left(14.0\mathrm{t}\right)\stackrel{Áåœ}{\mathrm{j}}\right)}{\mathrm{dt}} \\ =13.0\stackrel{Áåœ}{\mathrm{i}}+14.0\stackrel{Áåœ}{\mathrm{j}} \end{gathered}$$

Now, use the value of acceleration in the equation of Newton’s second law to calculate the force.

$$\begin{gathered} {\stackrel{â‡\P Ä}{\mathrm{F}}}_{\mathrm{net}}=\mathrm{m}\stackrel{â‡\P Ä}{\mathrm{a}} \\ -{\mathrm{F}}_1\cos {\mathrm{θ}}_1\stackrel{Áåœ}{\mathrm{i}}+{\mathrm{F}}_1\sin {\mathrm{θ}}_1\stackrel{Áåœ}{\mathrm{j}}+{\mathrm{F}}_2{\mathrm{²õ¾\pm ²Ôθ}}_2\stackrel{Áåœ}{\mathrm{i}}-{\mathrm{F}}_2{\mathrm{³¦´Ç²õθ}}_2\stackrel{Áåœ}{\mathrm{j}}+\stackrel{â‡\P Ä}{{\mathrm{F}}_3}=\mathrm{m}\left(13.0\stackrel{Áåœ}{\mathrm{i}}+14.0\stackrel{Áåœ}{\mathrm{j}}\right) \end{gathered}$$

Solve it for third force as,

$\stackrel{â‡\P Ä}{{\mathrm{F}}_3}=\mathrm{m}\left(13.0\stackrel{Áåœ}{\mathrm{i}}+14.0\stackrel{Áåœ}{\mathrm{j}}\right)+{\mathrm{F}}_1{\mathrm{³¦´Ç²õθ}}_1\stackrel{Áåœ}{\mathrm{i}}-{\mathrm{F}}_1{\mathrm{²õ¾\pm ²Ôθ}}_1\stackrel{Áåœ}{\mathrm{j}}-{\mathrm{F}}_2{\mathrm{²õ¾\pm ²Ôθ}}_2\stackrel{Áåœ}{\mathrm{i}}+{\mathrm{F}}_2{\mathrm{³¦´Ç²õθ}}_2\stackrel{Áåœ}{\mathrm{j}}$

Substitute the values of forces to calculate the third force.

$$\begin{gathered} \stackrel{â‡\P Ä}{{\mathrm{F}}_3}=0.0250\mathrm{kg}\left(13.0\stackrel{Áåœ}{\mathrm{i}}+14.0\stackrel{Áåœ}{\mathrm{j}}\right)\mathrm{m}/{\mathrm{s}}^2+\left(6\cos 30\right)\stackrel{Áåœ}{\mathrm{i}}-\left(6\sin 30\right)\stackrel{Áåœ}{\mathrm{j}}-\left(7\sin 30\right)\stackrel{Áåœ}{\mathrm{i}}+\left(7\cos 30\right)\stackrel{Áåœ}{\mathrm{j}} \\ =\left(2.02\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{i}}+\left(2.71\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{j}} \end{gathered}$$

Therefore, the third force is $\left(2.02\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{i}}+\left(2.71\mathrm{N}\right)\stackrel{Áåœ}{\mathrm{j}}$