Alright all of my forum friends that know physics.

I am stuck on this problem

I don't know where to begin since I havent seen a problem like this in my book homework . I just don't know how to attack it.

Draw a free-body diagram of the object and calculate the total x and y forces (in N) required to accomplish this acceleration. You will end up with the total required force vector.

If you then subtract the known force vectors (f1 and f2) from the total require force vector, you will find the remaining force vector (f3).

Make sense?

F1 doesn't have a y so I guess it is just zero?

Correct, F1 does not include a y contribution so it is just 0y.

Force Vectors are static problems, this looks like a motion problem.

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F = ma, or F=m(aX+aY)

They've given you m and aX + aY. Calculate the overall FX and FY.

Use the overall FX and FY, subtract out F1 and F2. The resultant components should define F3. Watch your signs and your units and it should be straightforward.

If it's computer graded (as appears) make sure you use three significant figures and a comma: F3x,F3y

Only if you are a civil engineer.

I wish I had seen that.. its so simple lol

Yeah, DUH!

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I got this

(-3.50 N, 0.523 N ) but it isn't accepting it as correct. which means I'm incorrect

Show your work?

Without crunching any numbers, it appears that the y value will need to be negative. Can you show what you did.

(also made another mistake, used 1.67 M/S^2 not 1.17)

sum of all x = 3.50N +(-1.20N)+ F3x = 1.17m/s^2

sum of all y = 0+ 1.78N + F3y = -0.677m/s^2

object has mass 6.17 kg (I think this might need to be used since the sum of both values is in Newtons and I guess everything has to be the same unit to be able to work with them?)

and I got it

answer is 4.92N ,-5.96 N

I had to get the sum of all forces for each by multiplying 6.17KG by the a for x and the same for y.. then the rest was elementary mathematics.

Your answer appears correct, but you're mixing units. Force (N) can't equal acceleration (m/s^2).

Fx = 3.50N +(-1.20N)+ F3x = (1.17m/s^2 )(6.17 kg)
Fy = 0+ 1.78N + F3y = (-0.677m/s^2)(6.17 kg)

F3x = 7.22 N - 3.5 N + 1.2 N = 4.92 N
F3y = -4.18 N - 1.78 N = -5.96

F=ma. If you know that, you can derive all other physics equations, almost.

Looks good.