Question

The resultant of two forces $PN$ and $3N$ is a force of $7N$. If the direction of $3N$ force were reversed, the resultant would be $\sqrt{19}N$. The value of $P$ is

• A. $5N$
• B. $6N$
• C. $3N$
• D. $4N$

Answer 1

The correct option is A

$5N$

Step 1: Given data

First force$\left(F_1\right)=PN$

Second force$\left(F_2\right)=3N$

Resultant force$\left(R\right)=7N$

Resultant force after reversing direction=$\sqrt{19}N$

Step 2: Formula used

The resultant between the two forces can be computed using the following formula:$R=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }WhereR=Resultantforce,F_1=firstforce,F_2=secondforce$

Step 3: Obtain the first equation

Consider the following formula,

Substitute the known values in the formula $R=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$,

$$\begin{gathered} 7=\sqrt{P^2+3^2+2\times P\times 3\cos \theta } \\ 7=\sqrt{P^2+9+6P\cos \theta } \\ 49=P^2+9+6P\cos \theta ·······\left(1\right) \end{gathered}$$

Step 4: Obtain the second equation

When the direction of $3N$ is reversed then $F_2$ becomes $-3N$ and $R$ become $\sqrt{19}N$. So,

Again substituting the known values in the formula gives,

$$\begin{gathered} \sqrt{19}=\sqrt{P^2+{\left(-3\right)}^2+2\times P\times -3\cos \theta } \\ \sqrt{19}=\sqrt{P^2+9-6P\cos \theta } \\ 19=P^2+9-6P\cos \theta ·······\left(2\right) \end{gathered}$$

Step 5: Find the value of $P$

Add the equation $\left(1\right)$ and $\left(2\right)$,

$$\begin{gathered} 49+19=P^2+9+6\cos \theta +P^2+9-6\cos \theta \\ 68=2P^2+18 \\ 50=2P^2 \\ 25=P^2 \\ P=\pm 5 \end{gathered}$$

From the figure, it is clear that the negative value can be ignored because the force P is acting in the upward direction.

Thus, the value of $P$ will be $5$.

Hence, option A is the correct answer.