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Problem : A 10 kg mass, initially at rest, experiences three forces: one North with magnitude 10 N, one East, with magnitude 20 N and one Northeast with magnitude 30 N. Find the resulting acceleration. After 10 seconds, assuming the forces continue to act while the object is in motion, what is the object's velocity? How far has it traveled?

We solve the problem by drawing a free body diagram:

Solution 1
Now we find the sum of the x and y components of all three forces:

Fx = 20 + 30 cos 45 = 41.2

Fy = 10 + 30 sin 45 = 31.2

And sum these two vectors. The magnitude of the resultant force is given by:

F = = = 51.7

And the direction is given by: θ = tan-1(31.2/41.2) = 37.1o, North of East. Thus the object experiences a force of 51.7 Newtons, directed 37.1o North of East. Now we must find its acceleration, using Newton's Second Law:

a = = = 5.17 , 37.1o    North of East

To find the final velocity and position of the object, we simply use the equations learned in kinematics:

vf = vo + at = 0 + (5.17)(10) = 51.7 m/s

xf = xo + vot + (1/2)at2 = 0 + 0 + (1/2)(5.17)(10)2 = 258.5 m

Thus, after 10 seconds, the object is moving with a velocity of 51.7 m/s, directed 31.7 degrees North of East. Also, the object has moved 258.5 meters, in the same direction.

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