Question

The following forces are acting on a particle (i) (2i^ + 3j^ – 2k^) N (ii) (3i^ + j^ – 3k^)N (iii) (-5i^ – 2j^ + k^)N the particle will move in * (a)x-y plane (b)x-z plane (c)y-z plane (d)aong x-axis​

Answer 1

Answer:

(c) y z plane

Explanation:

Add three vectors of force

=>2i +3j -2k +3i + j - 3k -5i -2j +k

=>0i +2j -4k

= 2j -4k N

Force is acting in Y and Z directions so the particle will move in y-z plane

Answer 2

First of all, lets find out the net force on the particle:

• The net force on the object is calculated by vector addition of the individual forces along all the axes.

$\rm \therefore \: F = F_{1} + F_{2} + F_{3}$

$\rm \implies F = (2\hat{i}+3\hat{j} - 2\hat{k}) + (3\hat{i}+\hat{j} - 3\hat{k}) + ( - 5\hat{i} - 2\hat{j}+\hat{k})$

$\rm \implies F = (2 + 3 - 5)\hat{i}+(3 + 1 - 2)\hat{j} + ( - 2 - 3 + 1)\hat{k}$

$\rm \implies F = 2\hat{j} - 4\hat{k}$

So, the object moves along the Y-Z plane.