a force is inclined to 50 degree to the horizontal. if its rectangular compound is horizontal direction be 50n find the magnitude of force and its vertical component ...
Answers
Answered by
102
Let the force be F
it is inclined at 50 degree with horizontal.
So F cos50 = 50N
⇒ F = 50/cos50 N
⇒ F = 77.78 N
vertical component, F sin50 = 77.78×sin50 = 59.58 N
it is inclined at 50 degree with horizontal.
So F cos50 = 50N
⇒ F = 50/cos50 N
⇒ F = 77.78 N
vertical component, F sin50 = 77.78×sin50 = 59.58 N
Answered by
35
Angle =50 degrees
Force = 50 N
So, Fcos50 = 50N
F=77.78N
Vertical component
= Fsin50
=77.78× sin50
=59.58N
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