a) Two men carry a weight 50N between two strings fixed to the weight, one string is inclined at 30^ to the vertical and the other at 60^ , find the tension of each string. [4] b) A body slides down…

a) Finding the tension in each string

Given:

• Total weight carried, $W=50\text{N}$
• Two strings inclined at angles $30^{\circ }$ and $60^{\circ }$ to the vertical.

Assumptions:

• The weight is in equilibrium (no acceleration).
• The strings are massless and inextensible.
• The system is symmetric about the vertical.

Approach: We resolve the tensions into vertical and horizontal components and apply equilibrium conditions.

Step 1: Define variables

Let:

• $T_1$ = tension in the string inclined at $30^{\circ }$ to the vertical.
• $T_2$ = tension in the string inclined at $60^{\circ }$ to the vertical.

Step 2: Resolve tensions into components

For $T_1$ (inclined at $30^{\circ }$ to the vertical):

• Vertical component: $T_1\cos 30^{\circ }$
• Horizontal component: $T_1\sin 30^{\circ }$

For $T_2$ (inclined at $60^{\circ }$ to the vertical):

• Vertical component: $T_2\cos 60^{\circ }$
• Horizontal component: $T_2\sin 60^{\circ }$

Step 3: Apply equilibrium conditions

Since the weight is in equilibrium:

1. Vertical equilibrium: $$T_1\cos 30^{\circ }+T_2\cos 60^{\circ }=W$$ Substituting $W=50\text{N}$: $$T_1\cos 30^{\circ }+T_2\cos 60^{\circ }=50\text{(1)}$$

2. Horizontal equilibrium: The horizontal components must cancel out: $$T_1\sin 30^{\circ }=T_2\sin 60^{\circ }\text{(2)}$$

Step 4: Solve equation (2) for $T_2$ in terms of $T_1$

$$T_1\sin 30^{\circ }=T_2\sin 60^{\circ }$$ $$T_2=T_1\frac{\sin 30^{\circ }}{\sin 60^{\circ }}$$ $$T_2=T_1\frac{0.5}{\sqrt{3}/2}=T_1\frac{1}{\sqrt{3}}=\frac{T_1}{\sqrt{3}}$$

Step 5: Substitute $T_2$ into equation (1)

$$T_1\cos 30^{\circ }+\left(\frac{T_1}{\sqrt{3}}\right)\cos 60^{\circ }=50$$ $$T_1\left(\frac{\sqrt{3}}{2}\right)+\frac{T_1}{\sqrt{3}}\left(\frac{1}{2}\right)=50$$ $$T_1\left(\frac{\sqrt{3}}{2}+\frac{1}{2\sqrt{3}}\right)=50$$ $$T_1\left(\frac{3+1}{2\sqrt{3}}\right)=50\text{(common denominator)}$$ $$T_1\left(\frac{4}{2\sqrt{3}}\right)=50$$ $$T_1\left(\frac{2}{\sqrt{3}}\right)=50$$ $$T_1=50\times \frac{\sqrt{3}}{2}=25\sqrt{3}\text{N}$$

Step 6: Find $T_2$

$$T_2=\frac{T_1}{\sqrt{3}}=\frac{25\sqrt{3}}{\sqrt{3}}=25\text{N}$$

Final Answer:

• Tension in the string inclined at $30^{\circ }$ to the vertical: $25\sqrt{3}\text{N}$ (≈ 43.3 N)
• Tension in the string inclined at $60^{\circ }$ to the vertical: $25\text{N}$

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b) Dividing the inclined plane into three equal-time intervals

Given:

• Height of the plane, $h=44.1\text{m}$
• Inclination angle, $\theta =30^{\circ }$
• Acceleration due to gravity, $g=9.8{\text{m/s}}^2$
• The body starts from rest.

Assumptions:

• The plane is smooth (no friction).
• The body slides under gravity only.

Step 1: Find the length of the inclined plane

Using trigonometry: $$h=L\sin \theta ⟹L=\frac{h}{\sin 30^{\circ }}=\frac{44.1}{0.5}=88.2\text{m}$$

Step 2: Find the acceleration along the plane

The component of gravity along the plane: $$a=g\sin \theta =9.8\times \sin 30^{\circ }=9.8\times 0.5=4.9{\text{m/s}}^2$$

Step 3: Find total time to slide down

Using the equation of motion: $$s=ut+\frac{1}{2}a{t}^2$$ Since $u=0$: $$88.2=0+\frac{1}{2}\times 4.9\times t^2$$ $$88.2=2.45t^2⟹t^2=\frac{88.2}{2.45}=36⟹t=6\text{s}$$

Step 4: Divide the total time into three equal intervals

Each interval: $\Delta t=\frac{6}{3}=2\text{s}$

Step 5: Find distances covered in each interval

We use the equation: $$s_n=ut+\frac{1}{2}a{t}^2$$ Since $u=0$, the distance covered in the $n$-th interval is: $$s_n=\frac{1}{2}a(t_n^2-t_{n-1}^2)$$ where $t_n=n\Delta t$.

First interval (0 to 2 s): $$s_1=\frac{1}{2}\times 4.9\times (2^2-0^2)=2.45\times 4=9.8\text{m}$$

Second interval (2 to 4 s): $$s_2=\frac{1}{2}\times 4.9\times (4^2-2^2)=2.45\times (16-4)=2.45\times 12=29.4\text{m}$$

Third interval (4 to 6 s): $$s_3=\frac{1}{2}\times 4.9\times (6^2-4^2)=2.45\times (36-16)=2.45\times 20=49\text{m}$$

Verification: Total distance covered: $$9.8+29.4+49=88.2\text{m}\text{(matches the length of the plane)}$$

Step 6: Find the heights corresponding to these distances

Since $h=L\sin \theta$, the height corresponding to a distance $s$ along the plane is: $$h_s=s\sin \theta$$

First division point (after 9.8 m): $$h_1=9.8\times \sin 30^{\circ }=9.8\times 0.5=4.9\text{m}$$

Second division point (after 9.8 + 29.4 = 39.2 m): $$h_2=39.2\times \sin 30^{\circ }=39.2\times 0.5=19.6\text{m}$$

Third division point (after 9.8 + 29.4 + 49 = 88.2 m): $$h_3=44.1\text{m}\text{(end of the plane)}$$

Final Answer: The inclined plane should be divided at heights of 4.9 m and 19.6 m from the bottom to ensure the body takes equal time intervals (2 seconds each) to traverse each segment.

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OR

Finding the depth of the well

Given:

• Total time for sound to travel back, $t=4\frac{2}{9}\text{s}=\frac{38}{9}\text{s}$
• Velocity of sound, $v_{\text{sound}}=352.8\text{m/s}$
• Acceleration due to gravity, $g=9.8{\text{m/s}}^2$

Assumptions:

• The stone is dropped from rest.
• Air resistance is negligible.
• The sound travels at a constant speed.

Step 1: Define variables

Let:

• $h$ = depth of the well.
• $t_1$ = time taken by the stone to fall to the water.
• $t_2$ = time taken by the sound to travel back to the top.

Step 2: Relate times

$$t_1+t_2=\frac{38}{9}\text{s}$$

Step 3: Express $t_1$ and $t_2$ in terms of $h$

For the stone falling: $$h=\frac{1}{2}g{t}_1^2⟹t_1=\sqrt{\frac{2h}{g}}$$

For the sound traveling back: $$t_2=\frac{h}{v_{\text{sound}}}$$

Step 4: Substitute and solve for $h$

$$\sqrt{\frac{2h}{g}}+\frac{h}{v_{\text{sound}}}=\frac{38}{9}$$ Let $h=x$: $$\sqrt{\frac{2x}{9.8}}+\frac{x}{352.8}=\frac{38}{9}$$ Simplify: $$\sqrt{\frac{x}{4.9}}+\frac{x}{352.8}=\frac{38}{9}$$

Let $y=\sqrt{x}$, then $x=y^2$: $$\frac{y}{\sqrt{4.9}}+\frac{y^2}{352.8}=\frac{38}{9}$$ $$\frac{y}{2.2136}+\frac{y^2}{352.8}=4.2222$$

Multiply through by $352.8\times 2.2136$ to eliminate denominators (approximate): $$160y+y^2=4.2222\times 352.8\times 2.2136$$ Calculate the right-hand side: $$4.2222\times 352.8\times 2.2136\approx 4.2222\times 782.4\approx 3300$$ So: $$y^2+160y-3300=0$$

Solve the quadratic equation: $$y=\frac{-160\pm \sqrt{160^2+4\times 3300}}{2}$$ $$y=\frac{-160\pm \sqrt{25600+13200}}{2}=\frac{-160\pm \sqrt{38800}}{2}$$ $$\sqrt{38800}\approx 197$$ $$y=\frac{-160+197}{2}=\frac{37}{2}=18.5\text{(discard negative root)}$$

Now, find $x$: $$x=y^2=(18.5{)}^2=342.25\text{m}$$

Verification: Calculate $t_1$ and $t_2$: $$t_1=\sqrt{\frac{2\times 342.25}{9.8}}=\sqrt{69.84}\approx 8.36\text{s}$$ $$t_2=\frac{342.25}{352.8}\approx 0.97\text{s}$$ Total time: $$8.36+0.97\approx 9.33\text{s}\text{(close to }\frac{38}{9}\approx 4.22\text{s?)}$$

Correction: There seems to be a miscalculation. Let’s re-solve the quadratic properly.

Rewriting the equation: $$\sqrt{\frac{2h}{9.8}}+\frac{h}{352.8}=\frac{38}{9}$$ Let $h=80\text{m}$ (trial): $$\sqrt{\frac{160}{9.8}}+\frac{80}{352.8}\approx 4.04+0.227\approx 4.267\text{s}\text{(close to }4.22\text{s)}$$

Let $h=78.4\text{m}$: $$\sqrt{\frac{156.8}{9.8}}+\frac{78.4}{352.8}\approx 3.98+0.222\approx 4.202\text{s}$$

Let $h=78.5\text{m}$: $$\sqrt{\frac{157}{9.8}}+\frac{78.5}{352.8}\approx 3.985+0.2225\approx 4.2075\text{s}$$

The correct depth is approximately 78.4 m.

Final Answer: The depth of the well is approximately 78.4 meters.