a) Two forces acting at an angle 45^ have a resultant equal to 10\ N . If one of the forces be 2\ N , find the other force. [2] b) Two parallel forces of 25 kg weight and 20 kg weight are acting at…

a) Finding the Other Force Given Resultant and Angle

Given:

• Two forces ${\mathbf{F}}_1$ and ${\mathbf{F}}_2$ act at an angle of $45^{\circ }$.
• Resultant force $\mathbf{R}=\sqrt{10}\mathrm{N}$.
• One force ${\mathbf{F}}_1=\sqrt{2}\mathrm{N}$.

To find: The magnitude of the other force ${\mathbf{F}}_2$.

Solution: The resultant of two forces acting at an angle $\theta$ is given by the formula: $$R=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$$ Here, $\theta =45^{\circ }$, $R=\sqrt{10}\mathrm{N}$, and $F_1=\sqrt{2}\mathrm{N}$.

Substitute the known values: $$\sqrt{10}=\sqrt{(\sqrt{2}{)}^2+F_2^2+2(\sqrt{2})F_2\cos 45^{\circ }}$$ Simplify: $$\sqrt{10}=\sqrt{2+F_2^2+2\sqrt{2}{F}_2\cdot \frac{\sqrt{2}}{2}}$$ $$\sqrt{10}=\sqrt{2+F_2^2+2F_2}$$ Square both sides: $$10=2+F_2^2+2F_2$$ Rearrange: $$F_2^2+2F_2-8=0$$ Solve the quadratic equation: $$F_2=\frac{-2\pm \sqrt{4+32}}{2}=\frac{-2\pm \sqrt{36}}{2}=\frac{-2\pm 6}{2}$$ Discard the negative solution (since force cannot be negative): $$F_2=\frac{4}{2}=2\mathrm{N}$$

Answer: The other force is 2 N.

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b) Finding the Point of Action of the Resultant of Two Parallel Forces

Given:

• Two parallel forces: $F_1=25\mathrm{kg}$ weight and $F_2=20\mathrm{kg}$ weight.
• Distance between the forces: $25\mathrm{cm}=0.25\mathrm{m}$.

To find: The point at which their resultant acts.

Solution: The resultant of two parallel forces acts at a point that divides the line joining their points of action in the ratio of their magnitudes (inverse ratio for weights).

Let the distance from the 25 kg weight to the resultant point be $x$. Then: $$\frac{x}{0.25-x}=\frac{20}{25}$$ Cross-multiply: $$25x=20(0.25-x)$$ $$25x=5-20x$$ $$45x=5$$ $$x=\frac{5}{45}=\frac{1}{9}\mathrm{m}\approx 11.11\mathrm{cm}$$

Answer: The resultant acts at a distance of 11.11 cm from the 25 kg weight.

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c) Finding the Velocity Acquired by a Body Under a Force

Given:

• Mass of the body $m=0.5\mathrm{kg}$.
• Force applied $F=2\mathrm{N}$.
• Time $t=1\sec$.
• Initial velocity $u=0m/s$.

To find: The velocity acquired during the second.

Solution: Using Newton’s Second Law: $$F=ma⟹a=\frac{F}{m}=\frac{2}{0.5}=4m/{\mathrm{s}}^2$$ Using the equation of motion: $$v=u+at$$ Since the body starts from rest ($u=0$): $$v=0+4\times 1=4m/s$$

Answer: The velocity acquired during the second is 4 m/s.

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Complete Answer (Structured for NEB Exam)

a) Finding the Other Force

Two forces ${\mathbf{F}}_1$ and ${\mathbf{F}}_2$ act at an angle of $45^{\circ }$, producing a resultant $\mathbf{R}=\sqrt{10}\mathrm{N}$. Given $F_1=\sqrt{2}\mathrm{N}$, we use the resultant formula: $$R=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos 45^{\circ }}$$ Substituting values: $$\sqrt{10}=\sqrt{2+F_2^2+2F_2}$$ Solving the quadratic equation yields: $$F_2=2\mathrm{N}$$

b) Finding the Resultant’s Point of Action

For two parallel forces of 25 kg and 20 kg acting 25 cm apart, the resultant divides the distance in the inverse ratio of their magnitudes: $$\frac{x}{0.25-x}=\frac{20}{25}$$ Solving gives: $$x=\frac{1}{9}\mathrm{m}\approx 11.11\mathrm{cm}$$

c) Finding the Velocity

A 0.5 kg body at rest is subjected to 2 N for 1 second. The acceleration is: $$a=\frac{F}{m}=4m/{\mathrm{s}}^2$$ The final velocity is: $$v=u+at=0+4\times 1=4m/s$$