a force is inclined at 30° to the horizontal.if its rectangular component in the horizontal direction is 50N ,find the magnitude of the force and its vertical components..??
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Answered by
3
Let the magnitude of force be F.
Fcos(30)=50
From here, F=100
Vertical component= Fcos(30)
=100 * sqrt(3)/2
=50*sqrt(3)
=50*1.7
=85 N
Fcos(30)=50
From here, F=100
Vertical component= Fcos(30)
=100 * sqrt(3)/2
=50*sqrt(3)
=50*1.7
=85 N
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Answered by
2
magnitude of force = F = ( Fx^2 + Fy^2 )^1/2
Fx = fcos (angle)
= 50*cos 30 =43.3
Fy = fsin (angle)
=50* sin 60
= 43.3
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