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A force is inclined at 60° to the horizontal.If its rectangular component in the horizontal direction is 50N, then magnitude of the force in the vertical direction is? (Plz give complete sol.)
one year ago

horizontal component of force = fcos60 = 50N ---------------------(1)
vertical component of force = fsin60
usinf equation 1 , we get f = 100N
now , put the value of f in vertical component
vertical component = 100 X 31/2/2 = 100 X 0.866 = 86.6 N
one year ago

Given that the force is inclined at 60° from the horizontal and  the horizontal component is 50.
Let the force be F . Therefore the force is given by.....
F x cos 60° = 50
F = 50 x 2
F = 100 units

The vertical component of the force is given by,
F x sin 60° =  vertical component
vertical component = 100 x √3/2
= 173.2/2
= 86.6 units

one year ago
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