The resultant of two forces 3P and 2P is R. I | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the angle between the two forces be $	heta$.Using the law of parallelogram of vectors,the resultant $R$ is given by:$R^2 = (3P)^2 + (2P)^2 +

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) Let the angle between the two forces be $	heta$.Using the law of parallelogram of vectors,the resultant $R$ is given by:$R^2 = (3P)^2 + (2P)^2 + 2(3P)(2P) \cos 	heta$$R^2 = 9P^2 + 4P^2 + 12P^2 \cos 	heta = 13P^2 + 12P^2 \cos 	heta$ ... $(i)$When the first force is doubled,it becomes $6P$. The new resultant is $2R$:$(2R)^2 = (6P)^2 + (2P)^2 + 2(6P)(2P) \cos 	heta$$4R^2 = 36P^2 + 4P^2 + 24P^2 \cos 	heta = 40P^2 + 24P^2 \cos 	heta$ ... (ii)Divide equation (ii) by $4$:$R^2 = 10P^2 + 6P^2 \cos 	heta$ ... (iii)Equating $(i)$ and (iii):$13P^2 + 12P^2 \cos 	heta = 10P^2 + 6P^2 \cos 	heta$$3P^2 = -6P^2 \cos 	heta$$\cos 	heta = -3/6 = -1/2$Therefore,$	heta = 120^\circ$.

The resultant of two forces $3P$ and $2P$ is $R$. If the first force is doubled,then the resultant is also doubled. The angle between the two forces is ........... $^o$.

Similar Questions

Prove the associative law of vector addition.

When vector $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ is subtracted from vector $\vec{B}$,it gives a vector equal to $2\hat{j}$. Then the magnitude of vector $\vec{B}$ will be:

If the direction of a vector $\overrightarrow{A}$ is reversed,find $\Delta \overrightarrow{A}$ and $|\Delta \overrightarrow{A}|$.

Find the resultant of three vectors $\overrightarrow{OA}$,$\overrightarrow{OB}$,and $\overrightarrow{OC}$ as shown in the figure. The radius of the circle is $R$.

Two forces $3\,N$ and $2\,N$ are at an angle $\theta$ such that the resultant is $R$. The first force is now increased to $6\,N$ and the resultant becomes $2R$. The value of $\theta$ is ....... $^\circ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module