An overhead view of a 12 kg tire that is to be pulled by three horizontal ropes. One rope's force (F1=50 N) is indicated. The forces from the other ropes are to be oriented such that the tire's acceleration magnitude a is least. What is that least a if (a) F2=30 N, F3=20 N; (b) F2=30 N, F3=10 N; and (c) F2=F3=30 N?
What is that least?
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Hey Sun -- a) 50 north cancelled by 30+20 south ...
b) 50 north offset by 30+10 south leaving 10N/12kg = 5/6 m/s/s North ...
c) 50 "noon" cancelled by "5pm & 7pm" 30's (the "peace" sign) ... Regards :)
Never mind my comment -- I misunderstood the question.
(a) Since F2+F3=50 N = F1, the two forces have to both act opposite to F1. The resultant force and acceleration will be zero.
(b) Since F2+F3=40 N < F1, the two forces still have to both act opposite to F1. Now the resultant force is 10 N and the acceleration 10 N/12 kg = 0.83 m/s², which is the least it can be.
(c) Since F2+F3=60 N > F1, the two forces should act at an angle θ relative to the direction opposite of F1, such that their resultant, 2(30)cos θ =50. This implies cos θ =5/6, or θ =33.6°. (That's 180-33.6=±146° relative to F1.) The resultant force and acceleration will be zero.