A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body

#### Solution 1

2 m/s^{2}, at an angle of 37° with a force of 8 N

Mass of the body, *m* = 5 kg

The given situation can be represented as follows:

The resultant of two forces is given as:

`R = sqrt((8)^2+(-6)^2) = sqrt(64+36) = 10N`

*θ* is the angle made by *R* with the force of 8 N

`:. theta = tan^(-1) ((-6)/8) = -36.87^@`

The negative sign indicates that *θ* is in the clockwise direction with respect to the force of magnitude 8 N.

As per Newton’s second law of motion, the acceleration (*a*) of the body is given as:

*F *= *ma*

*`:. a = F/m = 10/5 = 2 "m/s"^2`*

#### Solution 2

Here m = 5 kg

`F_1 = 8 N and F_2 = 6 N`

The resultant force on the body

`F =sqrt(F_1^2 + F_2^2) = sqrt(8^2 + 6^2 N)`

=> `F= sqrt(64 + 36)N` = 10 N

The acceleration `a = F/m`

`=> a = 10/5 = 2 ms^(-2)` in the same direction as the resultant force

The direction of acceleration

`tan beta = 6/8 = 3/4 = 0.75`

or `beta = tan^(-1)(0.75)`

= `37^@` with 8 N force