a) State and prove triangle of forces. [4] OR Forces 1N, 2N, 3N act at a point in direction parallel to the sides of an equilateral triangle taken in order. Find their resultant. b) A stone is…

a) State and Prove Triangle of Forces

Statement:

If three forces acting at a point are represented in magnitude and direction by the sides of a triangle taken in order, then the resultant of the three forces is zero. This is known as the Triangle of Forces.

Proof:

Let three forces $\vec{P},\vec{Q},\vec{R}$ act at a point $O$ such that they can be represented by the sides of a triangle $ABC$ taken in order.

1. Representation of Forces:

   • Let $\vec{P}$ be represented by $\overrightarrow{AB}$.
   • Let $\vec{Q}$ be represented by $\overrightarrow{BC}$.
   • Let $\vec{R}$ be represented by $\overrightarrow{CA}$.

2. Vector Addition:

   • The resultant of $\vec{P}$ and $\vec{Q}$ is: $$\vec{P}+\vec{Q}=\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$$
   • Now, add $\vec{R}$ (which is $\overrightarrow{CA}$): $$(\vec{P}+\vec{Q})+\vec{R}=\overrightarrow{AC}+\overrightarrow{CA}=\vec{0}$$
   • Thus, the resultant of all three forces is zero.

3. Conclusion:

   • Since the resultant is zero, the three forces are in equilibrium.
   • This proves the Triangle of Forces theorem.

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OR

Forces Acting at a Point in an Equilateral Triangle

Given:

• Three forces $1\text{N},2\text{N},3\text{N}$ act at a point in directions parallel to the sides of an equilateral triangle taken in order.

Solution:

1. Representation:

   • Let the forces be represented by vectors $\vec{A},\vec{B},\vec{C}$ along the sides of an equilateral triangle.
   • Since the triangle is equilateral, the angle between any two forces is $120^{\circ }$.

2. Resultant Calculation:

   • Let $\vec{A}=1\text{N}$ along the positive x-axis.
   • $\vec{B}=2\text{N}$ at $120^{\circ }$ from $\vec{A}$.
   • $\vec{C}=3\text{N}$ at $240^{\circ }$ from $\vec{A}$.

3. Resolve Forces into Components:

   • $\vec{A}=(1,0)$
   • $\vec{B}=(2\cos 120^{\circ },2\sin 120^{\circ })=(-1,\sqrt{3})$
   • $\vec{C}=(3\cos 240^{\circ },3\sin 240^{\circ })=(-1.5,-\sqrt{3})$

4. Sum of Components:

   • $x$-component: $$1+(-1)+(-1.5)=-1.5\text{N}$$
   • $y$-component: $$0+\sqrt{3}+(-\sqrt{3})=0\text{N}$$

5. Resultant Magnitude: $$R=\sqrt{(-1.5{)}^2+0^2}=1.5\text{N}$$

   • The resultant acts along the negative x-axis.

Final Answer: The resultant force is 1.5 N in the direction opposite to the 1 N force.

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b) Meeting Point of Two Stones

Given:

• A stone is dropped from a tower of height $200\text{m}$.
• Another stone is projected vertically upwards from the ground with $50\text{m/s}$.
• $g=9.8{\text{m/s}}^2$.

Solution:

1. Equations of Motion:

   • For the dropped stone (from rest): $$s_1=ut+\frac{1}{2}g{t}^2=0+\frac{1}{2}g{t}^2=4.9t^2$$
   • For the projected stone (upwards): $$s_2=ut-\frac{1}{2}g{t}^2=50t-4.9t^2$$

2. Condition for Meeting:

   • The sum of distances covered by both stones must equal the height of the tower: $$s_1+s_2=200$$ $$4.9t^2+50t-4.9t^2=200$$ $$50t=200⟹t=4\text{s}$$

3. Height at Meeting Point:

   • Substitute $t=4\text{s}$ into $s_1$: $$s_1=4.9(4{)}^2=78.4\text{m}$$
   • Thus, the stones meet at a height of 78.4 m from the ground after 4 seconds.

Final Answer: The stones meet after 4 seconds at a height of 78.4 m from the ground.