Four horizontal forces act on object. The strengths and directions of three of these forces are 1 N, -21 N, and 31 N. From the acceleration of the object it is known that the net force acting on it is 5 N. What is the strength and direction (sign) of the fourth of the individual (horizontal) forces?

Question

matt101 · Answer

Set up an equation to reflect the situation:

$$F_1+F_2+F_3+F_4=F_{net}$$
The question gives you values for 4 of the 5 variables - plug those in and solve for the unknown!

anonymous · Answer

Okay so would I add the 1,-21, and 31 together and subtract the 5 to move it over?

matt101 · Answer

Well almost - you would add the 1, -21, and 31 together and subtract all that from 5. You're just isolating F4, the unknown force.

anonymous · Answer

I'm getting 6 for that. Would that be right? Thinking it would be -6.

anonymous · Answer

@matt101

matt101 · Answer

Yes the answer is -6 N.

$$F_4=F_{net}-F_1-F_2-F_3$$$$F_4=5-1-(-21)-31$$$$F_4=-6$$

anonymous · Answer

Thanks!