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CBSE Sample Paper 2023-24 · Section E · 4 marks

Q37: Case study: tug of war with three forces on a ring

Sample 2023-24Ch 10 · Vector Algebra
The question

Read the following passage and answer the questions given below:

Teams A,B,CA, B, C went for playing a tug of war game. Teams A,B,CA, B, C have attached a rope to a metal ring and is trying to pull the ring into their own area.

Team AA pulls with force F1=6i^+0j^F_1 = 6\hat i + 0\hat j kN,

Team BB pulls with force F2=−4i^+4j^F_2 = -4\hat i + 4\hat j kN,

Team CC pulls with force F3=−3i^−3j^F_3 = -3\hat i - 3\hat j kN,

(i) What is the magnitude of the force of Team AA ? (1 mark)

(ii) Which team will win the game? (1 mark)

(iii) (a) Find the magnitude of the resultant force exerted by the teams. (2 marks)

OR

(iii) (b) In what direction is the ring getting pulled? (2 marks)

Exam answer

We have

∣F⃗1∣=62+02=6 kN|\vec F_1| = \sqrt{6^2 + 0^2} = 6 \text{ kN}
Why?

The magnitude (length) of a vector comes from Pythagoras:

∣ai^+bj^∣=a2+b2|a\hat i + b\hat j| = \sqrt{a^2 + b^2}

For F⃗1\vec F_1⁠, a=6a = 6 and b=0b = 0⁠:

62+02=36+0=36=6\begin{aligned} \sqrt{6^2 + 0^2} &= \sqrt{36 + 0} \\ &= \sqrt{36} = 6 \end{aligned}
∣F⃗2∣=(−4)2+42=32=42 kN\begin{aligned} |\vec F_2| &= \sqrt{(-4)^2 + 4^2} \\ &= \sqrt{32} = 4\sqrt2 \text{ kN} \end{aligned} ∣F⃗3∣=(−3)2+(−3)2=18=32 kN\begin{aligned} |\vec F_3| &= \sqrt{(-3)^2 + (-3)^2} \\ &= \sqrt{18} = 3\sqrt2 \text{ kN} \end{aligned}
Why?

A negative component squares to a positive number: (−4)2=16(-4)^2 = 16 and (−3)2=9(-3)^2 = 9⁠.

(−4)2+42=16+16=32(−3)2+(−3)2=9+9=18\begin{aligned} (-4)^2 + 4^2 &= 16 + 16 = 32 \\ (-3)^2 + (-3)^2 &= 9 + 9 = 18 \end{aligned}

Simplify the roots with a square factor: 32=16×2=42\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt2 and 18=9×2=32\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt2⁠.

(i) Magnitude of the force of Team AA = 6 kN.

(ii)

F⃗1+F⃗3=3i^−3j^=3(i^−j^)\vec F_1 + \vec F_3 = 3\hat i - 3\hat j = 3(\hat i - \hat j)
Why?

Add the i^\hat i parts and the j^\hat j parts separately: 6+(−3)=36 + (-3) = 3 and 0+(−3)=−30 + (-3) = -3⁠. This is the combined pull of Teams A and C.

(6i^+0j^)+(−3i^−3j^)=(6−3)i^+(0−3)j^=3i^−3j^=3(i^−j^)(take out 3)\begin{aligned} &(6\hat i + 0\hat j) + (-3\hat i - 3\hat j) \\ &= (6 - 3)\hat i + (0 - 3)\hat j \\ &= 3\hat i - 3\hat j \\ &= 3(\hat i - \hat j) \quad (\text{take out } 3) \end{aligned}
F⃗2=−4i^+4j^=−4(i^−j^)\vec F_2 = -4\hat i + 4\hat j = -4(\hat i - \hat j)

So, F⃗2\vec F_2 and F⃗1+F⃗3\vec F_1 + \vec F_3 are unlike vectors (opposite directions) having the same initial point, and ∣F⃗2∣=42|\vec F_2| = 4\sqrt2⁠, ∣F⃗1+F⃗3∣=32|\vec F_1 + \vec F_3| = 3\sqrt2⁠.

Why?

For F⃗2\vec F_2⁠, take out −4-4⁠: −4i^=−4(i^)-4\hat i = -4(\hat i) and 4j^=−4(−j^)4\hat j = -4(-\hat j)⁠, so

−4i^+4j^=−4(i^−j^)-4\hat i + 4\hat j = -4(\hat i - \hat j)

Both are multiples of the same vector i^−j^\hat i - \hat j⁠: one with a positive number (3) and one with a negative number (−4-4⁠). So they point in exactly opposite directions, like two sides of a tug of war.

∣F⃗1+F⃗3∣=3∣i^−j^∣=32|\vec F_1 + \vec F_3| = 3|\hat i - \hat j| = 3\sqrt2⁠, since ∣i^−j^∣=1+1=2|\hat i - \hat j| = \sqrt{1 + 1} = \sqrt2⁠.

Thus ∣F⃗2∣>∣F⃗1+F⃗3∣|\vec F_2| > |\vec F_1 + \vec F_3|⁠.

Why?

The stronger of two opposite pulls wins. 42>324\sqrt2 > 3\sqrt2⁠, so B's pull is larger. The net pull is

3(i^−j^)−4(i^−j^)=(3−4)(i^−j^)=−(i^−j^)\begin{aligned} &3(\hat i - \hat j) - 4(\hat i - \hat j) \\ &= (3 - 4)(\hat i - \hat j) \\ &= -(\hat i - \hat j) \end{aligned}

which points the way Team B pulls.

Hence, Team B will win the game.

(iii) (a)

F⃗=F⃗1+F⃗2+F⃗3=6i^+0j^−4i^+4j^−3i^−3j^=−i^+j^\begin{aligned} \vec F &= \vec F_1 + \vec F_2 + \vec F_3 \\ &= 6\hat i + 0\hat j - 4\hat i + 4\hat j - 3\hat i - 3\hat j \\ &= -\hat i + \hat j \end{aligned}
Why?

The resultant is the sum of all the forces. Collect like parts: i^\hat i⁠: 6−4−3=−16 - 4 - 3 = -1⁠; j^\hat j⁠: 0+4−3=10 + 4 - 3 = 1⁠.

i^:6−4=2,2−3=−1j^:0+4=4,4−3=1\begin{aligned} \hat i&: 6 - 4 = 2,\quad 2 - 3 = -1 \\ \hat j&: 0 + 4 = 4,\quad 4 - 3 = 1 \end{aligned}
∴ ∣F⃗∣=(−1)2+(1)2=2 kN\therefore\ |\vec F| = \sqrt{(-1)^2 + (1)^2} = \sqrt2 \text{ kN}
Why?

Same magnitude formula: 1+1=2\sqrt{1 + 1} = \sqrt2⁠. Keep the unit kN.

(−1)2+(1)2=1+1=2∣F⃗∣=2\begin{aligned} (-1)^2 + (1)^2 &= 1 + 1 = 2 \\ |\vec F| &= \sqrt2 \end{aligned}

The magnitude of the resultant force is 2\sqrt2 kN.

(iii) (b)

F⃗=−i^+j^\vec F = -\hat i + \hat j

Let θ\theta be the angle made by the resultant force with the positive direction of the xx-axis. Since F⃗\vec F has xx-component −1-1 and yy-component 11⁠,

θ=π−tan⁡−1(11)=π−π4=3π4\begin{aligned} \theta &= \pi - \tan^{-1}\left(\frac11\right) \\ &= \pi - \frac{\pi}{4} = \frac{3\pi}{4} \end{aligned}
Why?

The vector goes 1 unit left and 1 unit up, so it lies in the second quadrant. Its acute angle with the negative xx-axis has tan⁡=11=1\tan = \frac11 = 1⁠, so that angle is π4\frac{\pi}{4}⁠.

(tan⁡π4=1\tan\frac{\pi}{4} = 1⁠, so tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠.)

Measured from the positive xx-axis, the angle is π−π4\pi - \frac{\pi}{4}⁠.

π−π4=4π4−π4=4π−π4=3π4\begin{aligned} \pi - \frac{\pi}{4} &= \frac{4\pi}{4} - \frac{\pi}{4} \\ &= \frac{4\pi - \pi}{4} = \frac{3\pi}{4} \end{aligned}

Watch out: tan⁡−1(1−1)=−π4\tan^{-1}\left(\frac{1}{-1}\right) = -\frac{\pi}{4} points into the fourth quadrant, the wrong way.

The ring is pulled in the direction of −i^+j^-\hat i + \hat j⁠, at an angle 3π4\dfrac{3\pi}{4} with the positive direction of the xx-axis (the direction of Team B).

Where this comes from

Chapter 10, Vector Algebra: Vectors: magnitude, resultant and direction.

Quick lesson · Worked NCERT solutions · Interactive lesson

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