The two forces F 1 and F 2 shown in (Figure 1) act on a26.0-kg object on a frict

Question

The two forces F 1 and F 2 shown in (Figure 1) act on a26.0-kg object on a frictionless tabletop. Suppose thatF1 = 10.2 N , and F2 = 16.0 N .

Determine the magnitude of the net force on the object for the diagram (a) in the figure.

Determine the angle between the positive x axis and the net force F Ra on the object for the diagram (a), measured countercockwise.

Determine the magnitude of the acceleration of the object for the diagram (a).

Determine the magnitude of the net force on the object for the diagram (b) in the figure.

Determine the angle between the positive x axis and the net force F Rb on the object for the diagram (b), measured countercockwise.

Determine the magnitude of the acceleration of the object for the diagram (b).

Explanation / Answer

First off, I like to convert to "standard" angles -- measure counterclockwise (not

countercockwise") from the +x axis. so F1 is at 1 = 180º and F2 is at 2 = 270º.

Then the resultant R has components
Rx = F1*cos1 + F2*cos2 = 10.2N*cos180º + 16N*cos270º = -10.2 N
and
Ry = F1*sin1 + F2*sin2 = 10.2N*sin180º + 16N*sin270º = -16 N

Then the magnitude
|R| = (Rx² + Ry²) = 18.97 N

= arctan(-16 / -10.2) = 57.5º
and since Rx and Ry are negative, we know that the resultant is in quadrant III and the angle is 57.5º ccw from +x.

a = |R| / m = 18.97N / 26kg = 0.7296 m/s²

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