# Coplanar Forces acting on an object at different magnitudes and directions.

Three coplanar forces act on a 75000.0 kg object in outer space, where there is no gravity: 14.2 kN directed at 357Â°, 25.6 kN at 138Â°, and 16.4 kN at 255Â°.

What is the magnitude and direction of the acceleration?

If the object has an initial velocity of 17.5 miles/h directed at 58.6Â° in the same plane as all three forces, what will its velocity and position be after the force has been applied for 65.5 s?

What will the objects direction from the origin be and in what direction will it be moving assuming the force was applied at the origin of coordinates? All directions are with respect to the x direction.

Please show all work.

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#### Solution Preview

Force vectors:

vF1(read as F1 vector) = 14.2 *(i*cos(357) + j*sin(357))

=>vF1 = 14.1805i - 0.7432j

vF2 = 25.6*(i*cos(138) + j*sin(138)) = -19.0245i + 17.1297j

vF3 = 16.4*(i*cos(255) + j*sin(255)) = - 4.2446i -15.8412j

net force:

vF = vF1+vF2+vF3= (14.1805-19.0245-4.2446)i +(-0.7432+17.1297-15.8412)j

=> vF = -9.0886i + 0.5453j

Note: the magnitude forces are in kN

therefore ...

#### Solution Summary

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