Question

A particle is simultaneously acted upon by two forces of magnitude $3N$ and $4N$. The net force on the particle is:
A. $7N$
B. $5N$
C. $1N$
D. Between $1N$ and $7N$

Answer 1

Hint: Force is a vector quantity. Vector quantities are those which have magnitude as well as direction. The addition of vector quantities is not like that of scalar quantities. There are some laws to add the vectors. To find the resultant of vectors along with their magnitude we also need the angle they make with each other. This concept we will discuss here.


Complete step by step solution:
To add two vectors, we have a vector addition law which is given below.
Suppose, we have two vectors A and B making angles $\theta $ with each other. Their results are given below.
$R = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } $
Here, two forces $3N$and $4N$ are acting on a particle, but the angle between them is not given. So, let us consider, either they can be acting in the same direction or opposite direction, that is, either $0$ or \[180\] degrees.
Let us first calculate the resultant force when both are in the same direction.
$R = \sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + 2 \times 3 \times 4 \times \cos 0} = \sqrt {9 + 16 + 24} $
Further simplifying.
$R = \sqrt {49} = 7N$
Let us now calculate the resultant force when both are in opposite directions.
$R = \sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + 2 \times 3 \times 4\cos 180} + \sqrt {9 + 16 - 24} $
Further simplifying.
$R = \sqrt 1 = 1N$
Therefore, resultant force can lie between $1N$ and $7N$ as all the angles between $0$ or \[180\] degrees are possible.
Hence, option (D) between $1N$ and $7N$ is correct.

Note:
Remember that for $0$ degrees and \[180\] degrees the resultant varies from (A+B) to (A-B).
Multiplication of vectors is different from the multiplication of a scalar quantity.
Multiplication of vectors is either defined by scalar product or by vector product.
Scalar product: $\vec A.\vec B = AB\cos \theta $
Vector product: $\vec A \times \vec B = AB\sin \theta \hat n$
Here, $\hat n$ is the direction of the unit normal.