Question

A force inclined at 60° to the horizontal. If its component in the horizontal direction is 50N, then the
magnitude of the force in the vertical direction is:​

Answer 1

Answer:

86.6 N

Explanation:

Horizontal component of Force = 50N

⇒ Fcos 60°= 50

⇒ F= 50/cos 60

⇒ F= 50/0.5= 100 N

Vertical component of force = Fsin60°= 100×0.866= 86.6 N

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