When the angle between force & displacement is less than 90 work done is positive because one component of force will be positive Fd cos theta i.e.Fdcos0=Fd×1=Fd & the other Fdsin theta is zero, but why it’s zero as sin90=1?
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Work is defined as F.d or |Fd| cosα where α is the angle between force and displacement.
If you want to find the work component wise and add, the two components of force would be F cosα (along the displacement) and F sinα (perpendicular to the displacement)
Now, component-1:Fcosα
angle between (F cosα) and the displacement is 0.
So W = (F cosα)×d× (cos 0) =Fd cosα × 1 = Fd cosα (not Fd as you have written)
component-2:Fsinα
angle between (F sinα) and displacement is 90.
So W = (F sinα)×d×(cos 90) = Fd sinα × 0 = 0 (not Fd sinα)
If you want to find the work component wise and add, the two components of force would be F cosα (along the displacement) and F sinα (perpendicular to the displacement)
Now, component-1:Fcosα
angle between (F cosα) and the displacement is 0.
So W = (F cosα)×d× (cos 0) =Fd cosα × 1 = Fd cosα (not Fd as you have written)
component-2:Fsinα
angle between (F sinα) and displacement is 90.
So W = (F sinα)×d×(cos 90) = Fd sinα × 0 = 0 (not Fd sinα)
TPS:
if you still don't understand, leave a comment.
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