A 200 N force is applied to an object (that is at the origin) at 30 degrees above the horizontal on the positive x axis. A second force is applied at 20 degrees below the horizontal on the negative x axis. In order for the object to be in equilibrium in the y direction, what must the magnitude of this force be?

Question

Physics

Tori Ingram

3 November, 12:49

A 200 N force is applied to an object (that is at the origin) at 30 degrees above the horizontal on the positive x axis. A second force is applied at 20 degrees below the horizontal on the negative x axis. In order for the object to be in equilibrium in the y direction, what must the magnitude of this force be?

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Answers (1)

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Jacob Cannon · Answer 1

294.11 N

Explanation:

F1 = 200 N

Let the other force is F2 = F = ?

Resolve the components of F1 and F2.

As the object is in equilibrium in y direction, it means the net force in y direction is zero.

So, F1 Sin 30 = F2 Sin 20

200 x 0.5 = F x 0.34

F = 294.11 N

The magnitude of force is 294.11 N