Question

What is the force at C point, mass of the object $= 1{\text{ kg?}}$

$A.){\text{ 2}}{\text{.5}} \\\ B.{\text{) 5}} \\\ C.{\text{) }}\dfrac{{5\sqrt 3 }}{2} \\\ D.){\text{ 0}} \\\$

Answer 1

Here, use Newton's second law of equation and place the given values and find the effective force. Along with the formula, use the derivative function and its application and simplify.

Complete step by step answer: By using Newton’s second law statement which states that the net external force action on the object is always equal to the rate of change of momentum with respect to the rate of change of the time.
It can be expressed as –
$F = \dfrac{{dp}}{{dt}}$
The above equation can be written as –
$F = \dfrac{{dp}}{{dt}} \cdot \dfrac{{dx}}{{dx}}$
It can be re-written as –
$F = \dfrac{{dp}}{{dx}} \cdot \dfrac{{dx}}{{dt}}$
Now, by the definition of the velocity – it is distance per unit time.
$\Rightarrow F = \dfrac{{dp}}{{dx}} \cdot v$ ..... (a)
By referring diagram,
$\sin 30^\circ = \dfrac{v}{5}$
Place the value $\sin 30^\circ = \dfrac{1}{2}$ in the above equation –
$\dfrac{1}{2} = \dfrac{v}{5}$
Do cross-multiplication and make the velocity “v” the subject in the above equation –
$v = \dfrac{5}{2}m/s$ ....... (b)
Now, change of momentum with respect to distance is
$\dfrac{{dp}}{{dx}} = \tan 60^\circ$
(Since sum of vertical opposite angle and the other angle make $90^\circ$, so here we get $\theta = 60^\circ$ )
Place the value of $\tan 60^\circ = \sqrt 3$ in the above equation –
$\dfrac{{dp}}{{dx}} = \sqrt 3$ ....... (c)
Place the value of the equations (b) and (C) in the equation (a)

$$F = \dfrac{5}{2} \cdot \sqrt 3 \\\ \Rightarrow F = \dfrac{{5\sqrt 3 }}{2}N \\\$$

Hence, option C is the correct answer.

Note: To solve these types of sums, remember the basic three Newton’s laws of motion and its formula along with its correct application. Remember the basic trigonometric values for sine, cosine and tangent functions for all the angles at least for $0^\circ ,30^\circ ,45^\circ ,60^\circ ,90^\circ$ for direct substitution.