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Question
A body of mass 10 kg is acted upon by two perpendicular forces, 6 N and 8 N. The resultant acceleration of the body is ______.
- 1 m s–2 at an angle of tan−1 `(4/3)` w.r.t 6 N force.
- 0.2 m s–2 at an angle of tan−1 `(4/3)` w.r.t 6 N force.
- 1 m s–2 at an angle of tan−1 `(3/4)` w.r.t 8 N force.
- 0.2 m s–2 at an angle of tan−1 `(3/4)` w.r.t 8 N force.
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Solution
a and c
Explanation:
Consider the adjacent diagram
Given, mass = m = 10 kg
F1 = 6 N, F2 = 8 N
Resultant force = F = `sqrt(F_1^2 + F_2^2)`
= `sqrt(36 + 64)`
= 10 N
a = `F/m = 10/10` = 1 m/s2; along R.
Let θ1 be the angle between R and F1.
tan θ1 = `8/6 = 4/3`
θ1 = `tan^-1 (4/3)` w.r.t. F1 = 6 N
Let θ2 be the angle between F and F2.
tan θ2 = `6/8 = 3/4`
θ2 = `tan^-1 (3/4)` w.r.t. F2 = 8 N
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