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Motion In A Plane

Question
CBSEENPH11018249

A force inclined at angle 37° with the horizontal has vertical component 42N. Find the magnitude of force and magnitude of its horizontal component. (Take sin 37° = 0·6 and cos 37° = 0·8)

Solution
Let F be the magnitude of force.
If force vector subtends an angle θ with the horizontal, then horizontal and vertical components of force respectively, 
              space space space straight F subscript straight x equals space Fcosθ     and     straight F subscript straight y space equals space Fsinθ
Here,
Angle with which the force is inclined to the horizontal, straight theta equals 37 degree
Vertical component of the force, straight F subscript straight y equals 42 space straight N 
So, magnitude of force is found using, 
∴               42 space equals space straight F space sin 37 degree space equals space straight F space cross times space 0.6
Force, straight F space equals space 70 straight N
Horizontal component, straight F subscript straight x equals Fcosθ space equals space 70 cross times 0.8 space equals space 56 straight N