Two vectors, A → & B → have the same magnitude. Under what circumstances does the vector A → + B → has the same magnitude as | A → | or | B → | . When does the vector difference A → - B → have this magnitude?

It is given that A → = B → = a (assume). Let, R → = A → + B → and S → = A → - B → . According to law of cosines, R → = A 2 + B 2 + 2 A B cos θ . For R → = A → + B → = a , we have, a = a 2 + a 2 + 2 a 2 cos θ ⇒ a 2 = 2 a 2 + 2 a 2 cos θ ⇒ a 2 = 2 a 2 ( 1 + cos θ ) ⇒ cos θ = 1 2 - 1 = - 1 2 ⇒ θ = cos - 1 - 1 2 = 120 ∘ . So, for A → = B → = A → + B → , angle between A → and B → must be 120 ∘ . For, S → = A → - B → = a , we have, a = a 2 + a 2 - 2 a 2 cos θ ⇒ a 2 = 2 a 2 - 2 a 2 cos θ = 2 a 2 ( 1 - cos θ ) ⇒ cos θ = 1 - 1 2 = 1 2 ⇒ θ = cos - 1 1 2 = 60 ∘ . So, for A → = B → = A → - B → , angle between A → and B → must be 60 ∘ .

Four ants 1 , 2 , 3 and 4 are pulling a grain with force of magnitudes 3 N , 1 N , 2 N and | F → | N as shown in the figure. Find the force, F → if, the grain remains in equilibrium under the action of the above forces.

Forces by ants have been shown in the following diagram. Let, the force, F → by the fourth ant be making an angle of ϕ with the negative direction of x - axis. Force by ant 1 has a magnitude of 3 N and makes an angle of 30 ° with the x - axis. Components of force by ant 1 are, 3 cos 30 ° = 3 3 2 N along positive x - axis and 3 sin 30 ° = 3 2 N along y - axis. Force by ant 2 is 1 N along the negative x - axis. Force by ant 3 makes an angle of - 45 ° with the x - axis. Hence, components of it are, 2 cos 45 ° N = 2 2 N = 2 N along positive x - axis and 2 sin 45 ° N = 2 2 N = 2 N along negative x - axis. Components of force by ant 4 are F cos ϕ along negative x - axis and F sin ϕ along negative y - axis. Since, the grain is in equilibrium, net force on it will be zero. Net force along x -axis is zero, so, ⇒ - 1 N - F cos ϕ + 2 N + 3 3 2 N = 0 ⇒ F cos ϕ = 3 3 2 + 2 - 1 N . . . ( 1 ) Net force along y -axis is zero, so, ⇒ 3 2 N - F sin ϕ - 2 N = 0 ⇒ F sin ϕ = 3 2 - 2 N . . . ( 2 ) Squaring equations 1 & 2 and adding them, we get, ⇒ F 2 cos 2 ϕ + F 2 sin 2 ϕ = 3 3 2 + 2 - 1 2 N 2 + 3 2 - 2 2 N 2 ⇒ F 2 cos 2 ϕ + sin 2 ϕ = F 2 = 3 3 2 + 2 - 1 2 N 2 + 3 2 - 2 2 N 2 ⇒ F = 3 3 2 + 2 - 1 2 + 3 2 - 2 2 N . Dividing equation 2 by 1 , we get, F sin ϕ F cos ϕ = 3 2 - 2 3 3 2 + 2 - 1 ⇒ tan ϕ = 3 - 2 2 3 2 + 2 2 - 2 ⇒ ϕ = tan - 1 3 - 2 2 3 2 + 2 2 - 2 .

The vector equation of a line which is parallel to a given vector A → and passes through a given point P having position vector r 0 → is,

Let r 0 → be the position vector of point P w.r.t. origin in the rectangular coordinate system, and let l be the line passing through point P and is parallel to A → . Let r → be the position vector of arbitrary point Q as shown in the figure. So, P Q → is parallel to A → . So, P Q → = λ A → , where λ is a real constant. Also, A → = A . A ^ . So, P Q → = λ A . A ^ = n A ^ , where n = λ A . But, by triangle law of vector addition, O P → + P Q → = O Q → ⇒ P Q → = O Q → - O P → ⇒ n A ^ = r → - r 0 → ⇒ r → = r 0 → + n A ^

A cylinder of radius, R is moving down with speed, v into water, placed in a cylinder of radius, 4 R as shown. Find the rate by which, (a) water surface is rising. (b) wet surface area of cylinder of radius, R is increasing.

(a) Given that, a cylinder of radius, R is moving down with a velocity, v into the water in a cylinder of radius, 4 R . The rate of increase of the submerged length, x of the moving cylinder with respect to the fixed cylinder is d x d t = v . . . 1 Let, h be the rate at which water rises in the fixed cylinder. Then water displaced by the moving cylinder equals the volume due to the rise of the height of water. Rate of volume displaced by the cylinder is d V d t = d π R 2 x d t = π R 2 d x d t ⇒ d V d t = π R 2 v . . . 2 Rise of volume of water in the annular area is, V = h π 4 R 2 - π R 2 . Differentiating the above equation, the rate of rise of volume of the water due to rise in height is, d V d t = d h π 4 R 2 - π R 2 d t = 15 π R 2 d h d t . . . 3 Equating equations 2 and 3 , π R 2 v = 15 π R 2 d h d t ⇒ d h d t = π R 2 v 15 π R 2 = v 15 ⇒ d h d t = v 15 . . . ( 4 ) (b) Wet surface area of the moving cylinder is A = π R 2 + 2 π R ( x + h ) . So, the rate of increase of the wet area is d A d t = d π R 2 + 2 π R x + h d t = 2 π R d x d t + d h d t . Substituting equation ( 1 ) and 4 , d A d t = 2 π R d x d t + d h d t = 2 π R v + v 15 = 32 π R v 15 . Therefore, the rate of increase of the wet area of the cylinder of radius, R is 32 π R v 15 .

Embibe Experts Solutions for Chapter: Mathematical Tools in Physics, Exercise 3: Exercise - 4

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Attempt the free practice questions on Chapter 2: Mathematical Tools in Physics, Exercise 3: Exercise - 4 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Physics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Mathematical Tools in Physics, Exercise 3: Exercise - 4 with Hints & Solutions

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Alpha Question Bank for Engineering: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise - 4>Q 1

Two vectors $\vec{A}$ and $\vec{B}$ are defined as $\vec{A} = α \hat{i}$ and $\vec{B} = α (\cos ω t \hat{i} + \sin ω t \hat{j}),$ where $α$ is a constant and $ω = \frac{π}{6} rad s^{- 1} .$ If $| \vec{A} + \vec{B} | = \sqrt{3} | \vec{A} - \vec{B} |$ at time $t = τ$ for the first time, the value of $τ$ in second is _____.

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Alpha Question Bank for Engineering: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise - 4>Q 15

Find an antiderivative for the function. Do as many as you can mentally. Check your answer by differentiation.
$- πcosec \frac{π x}{2} \cot \frac{π x}{2}$

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Alpha Question Bank for Engineering: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise - 4>Q 49

Two vectors, $\vec{A} & \vec{B}$ have the same magnitude. Under what circumstances does the vector $\vec{A} + \vec{B}$ has the same magnitude as $| \vec{A} |$ or $| \vec{B} |$ . When does the vector difference $\vec{A} - \vec{B}$ have this magnitude?

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Alpha Question Bank for Engineering: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise - 4>Q 51

If five consecutive sides of a regular hexagon represent five unit vectors acting in the same sense, Their resultant vector is,

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Alpha Question Bank for Engineering: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise - 4>Q 52

Four ants $1, 2, 3$ and $4$ are pulling a grain with force of magnitudes $3 N, 1 N, 2 N$ and $| \vec{F} | N$ as shown in the figure. Find the force, $\vec{F}$ if, the grain remains in equilibrium under the action of the above forces.

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Alpha Question Bank for Engineering: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise - 4>Q 53

The vector equation of a line which is parallel to a given vector $\vec{A}$ and passes through a given point $P$ having position vector $\vec{r_{0}}$ is,

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Alpha Question Bank for Engineering: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise - 4>Q 56

A cylinder of radius, $R$ is moving down with speed, $v$ into water, placed in a cylinder of radius, $4 R$ as shown. Find the rate by which,

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(a) water surface is rising.

(b) wet surface area of cylinder of radius, $R$ is increasing.

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Alpha Question Bank for Engineering: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise - 4>Q 57

Maximum area of the rectangle that can be inscribed in a circle of radius $r$ is $m r^{2}$ . Find $m$ .

Embibe Experts Solutions for Chapter: Mathematical Tools in Physics, Exercise 3: Exercise - 4

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Questions from Embibe Experts Solutions for Chapter: Mathematical Tools in Physics, Exercise 3: Exercise - 4 with Hints & Solutions

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