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/s2, 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
