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This is second of our many quizzes on Force vectors. Click on this to access more such competitive quizzes on Mechanics.

Force vectors Quiz 2

physics

Suman applied force of $\left( 3\hat{i}+5\hat{j} \right)N$. Priyanka applied force of $\left( 9\hat{i}+7\hat{j} \right)\,N$ and Ananya applied force of $\left( 7\hat{i}+19\hat{j} \right)N$ on an almirah. What is the resultant force which acts on the almirah ? 

Begin by determining the net force exerted by each person on the almirah. Then find the resultant force, following which you can choose the option closest to the answer you get.  

  $\text{The net force applied by Suman will be:}\,F_{s} = \sqrt{3^2 + 5^2} = \sqrt{34} = 5.83\,N\\ \text{The net force applied by Priyanka will be:}\,F_{p} = \sqrt{9^2 + 7^2} = \sqrt{130} = 11.40\,N\\ \text{The net force applied by Ananya will be:}\,F_{a} = \sqrt{7^2 + 19^2} = \sqrt{410} = 20.25\,N\\ \text{Assuming that they all apply the force on the almirah in the same direction,}\\ \text{the resultant force acting on it will be:}\\ F_{r} = F_s + F_p + F_a\\ \Rightarrow F_{r} = 5.83 + 11.40 + 20.2\\ \Rightarrow F_{r} = 37.48\\ \therefore F_{r} = \sqrt{1405}\,N\\ \text{From the given options, the answer we get is closest to}\,\sqrt{24 \times 55} = \sqrt{1320}$

Ahem, Jigar and Neha are pushing a carton of mass $18\,Kg$ by applying forces ${{F}_{1}}=\left( 2\hat{i} \right)\,N$, ${{F}_{2}}=(6\hat{j})\,N$ and ${{F}_{3}}=\left( 9\hat{k} \right)\,N$ respectively. The resultant of all the forces is $R$. Now if Jigar stops pushing the carton, the resultant becomes ${{R}^{''}}$. What is the value of ${{\left( \dfrac{R}{{{R}^{''}}} \right)}^{2}}$ 

First find the net force exerted by each person, and then additively find the resultant. Keep in mind to discount the force applied by Jigar in the second case. 

 $\text{Net force exerted by Ahem:}\,F_{1} = \sqrt{2^2+0^2+0^2} = 2\,N\\ \text{Net force exerted by Jigar:}\,F_2 = \sqrt{0^2+6^2 + 0^2} = 6\,N\\ \text{Net force exerted by Neha:}\,F_3 = \sqrt{0^2 + 0^2+ 9^2} = 9\,N\\ \text{The resultant force R exerted by all three can be given as:}\\ R = F_1 + F_2 + F_3\\ \Rightarrow R = 2 +6+9\\ \Rightarrow R = 17\\ \text{The resultant force}\,R^{\prime \prime}\,\text{when Jigar stops pushing the carton will be:}\\ R^{\prime \prime} = F_1 + F_3\\ \Rightarrow R^{\prime \prime} = 2 + 9\\ \Rightarrow R^{\prime \prime} = 11\\ \text{Therefore}\,\left(\dfrac{R}{R^{\prime \prime}}\right)^2 = \left(\dfrac{17}{11}\right)^2 = \dfrac{289}{121}$ 

 Among the following pairs of the force vectors which pair of forces are inclined perpendicular to each other. 

The angle between the two vectors can be determined from the dot product of the two by analyzing the cosine value obtained. If the angle between them is $\,90^{\circ}$, then the pair of vectors are perpendicular.  

$\text{We can find the inclination between any two force vectors from their scalar product, i.e.,}\\ F_1 . F_2 = |F_1||F_2|\cos\theta\\ \Rightarrow \theta = cos^{-1}\left(\dfrac{F_1 .F_2}{|F_1||F_2|}\right)\\ \text{Checking for the first option:}\\ F_1 = \hat{i}-4\hat{j}+5\hat{k}\,N\,\text{and}\,F_2 = 2\hat{i}+3\hat{j}+2\hat{k}\\ \Rightarrow  |F_1| = \sqrt{1^2 + (-4^2)+ 5^2}\\ \Rightarrow  |F_1|= \sqrt{42}\\ \Rightarrow  |F_1|=6.48\,N\\ \text{And}\\ |F_2| = \sqrt{2^2 + 3^2 + 2^2}\\ \Rightarrow|F_2|= \sqrt{144}\\ \Rightarrow|F_2|= 12\,N\\ \text{Plugging these into the equation for}\, \theta:\\ \theta = cos^{-1}\left(\dfrac{(\hat{i}-4\hat{j}+5\hat{k}).(2\hat{i}+3\hat{j}+2\hat{k})}{6.48 \times 12}\right)\\ \Rightarrow\theta = cos^{-1}\left(\dfrac{2-12+10}{77.76}\right)\\ \therefore \theta = cos^{-1}(0) = 90\,^{\circ}\\ \text{Thus, the two forces are inclined perpendicular to each other.}$

 In a dream Arjun saw that he applied a force on a large trunk displacing it from $\left( 2\hat{i}+3\hat{j}+9\hat{k} \right)\,m$ to $\left( 7\hat{i}+9\hat{j}+19\hat{k} \right)\,m$. In this process he did a work of $200\,J$. Find out the velocity of the trunk in $m/s$ at t= $3s$ assuming its mass to be $4\,Kg$. 

First find the displacement from the position vectors and then determine the force applied to push the trunk through the given displacement, following which you can arrive at its velocity.  

$\text{Given that the trunk is displaced from}\,x_1 = 2\hat{i} + 3\hat{j}+9\hat{k}\\ \text{and}\,x_2 = 7\hat{i} + 9\hat{j}+19\hat{k}\,\text{the displacement of the trunk will be:}\\ s = x_2 – x_1\\ \Rightarrow s = (7-2)\hat{i} + (9-3)\hat{j} + (19-9)\hat{k}\\ \Rightarrow s = 5\hat{i} + 6\hat{j}+10\hat{k}\\ \text{The resultant displacement will be:}\\ s_{r} = \sqrt{5^2 + 6^2 + 10^2}\\ \Rightarrow s_{r} = \sqrt{161}\\ \Rightarrow s_{r} = 12.69\,m\\ \text{Now, if F is the force applied on the trunk,}\\ \text{the work done to push it through a displacement of}\,s_r\,\text{will be:}\\ W = F \times s_r\\ \Rightarrow F = \dfrac{W}{s_r}\\ \text{Given that}\,W = 200\,J\,\text{the above expression becomes:}\\ F = \dfrac{200}{12.69}\\ \Rightarrow F = 15.76\,N\\ \text{Now, the mass of the trunk is m = 4 kg.}\\ \text{Assuming that the trunk is pushed from rest, i.e., u = 0:}\\ F = ma\\ \Rightarrow F = m \times \dfrac{v-u}{t}\\ \Rightarrow F = \dfrac{mv}{t}\\ \Rightarrow v = \dfrac{F \times t}{m}\\ \Rightarrow v = \dfrac{15.76 \times 3}{4}\\ \therefore v = 11.8\,ms^{-1}$

A disco dancer swings a disco ball hanging from a thread attached to the ceiling of the room.The ball starts moving in a circular path with the velocity of $5\,m/s$. The force which the dancer applied on the ball is given as $F=\left( 4\hat{i}+3\hat{j} \right)N$ and mass of the ball is $13\,Kg$. What is the area of the circular path on which the ball moves? 

Recall that the force acting on a body executing circular motion will be the centripetal force. From this, determine the radius and consequently the area of the circular path traced by the ball. 

$\text{The force acting on the ball will be a centripetal force acting on it by virtue of its circular motion, given by:}\\ F_c = \dfrac{mv^2}{r}\\ \Rightarrow r = \dfrac{mv^2}{F_c}\\ \text{Now, given that}\,F=\left( 4\hat{i}+3\hat{j} \right)N\\ \text{The magnitude of this force acting on the ball will be:}\\ F_c = \sqrt{4^2 + 3^2}\\ \Rightarrow F_c = \sqrt{25}\\ \Rightarrow F_c = 5\,N\\ \text{Plugging in}\,v = 5\,ms^{-1}\,\text{and}\,m = 13\,kg\,\text{into the radius equation, we get:}\\ r = \dfrac{13 \times 5^2}{5}\\ \Rightarrow r = 65\,m\\ \text{Now, the area of the circular path traced by the ball will be:}\\ A= \pi r^2\\ \Rightarrow A = \pi \times 65^2\\ \therefore A = (\pi \times 4225)\,m^2$