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

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

What is the magnitude and direction of the acceleration?

If the object has an initial velocity of 17.5 miles/h directed at 58.6° 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.

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The very detailed solution provides explanations, formulas and calculations to arrive at the answers.

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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 ...

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• MSc , Pune University, India
• PhD (IP), Pune University, India
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