A vector representing 160 N is oriented at 51◦ with the horizontal. What is the magnitude of its horizontal component?

horizontal-component

#1

A vector representing 160 N is oriented at 51◦with the horizontal.What is the magnitude of its horizontal component?

Answer:

To get the horizontal component you need to use the formula F cos(theta), where F is the magnitude of the vector. So, in your case it will be 160 cos(51 deg) = 100.7 N

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