Question

A rogue chipmunk has been raiding the secrete acorn stash of 3 squirrels. On discovering the perpetrator in the very act they secure him with 3 ropes and attempt to pull him off the limb where he falls to his demise. The first squirrel pulls at an angle of 12 degrees with a force of .63 N. Squirrel two pulls at an angle of 158 and at a force of .48 N. The third squirrel at an angle of 93, and 1.15 N.
A) If the chipmunk is able to dig and resist them to a standstill, what is his equilibrent force?

Answer 1

Fs = 63N @ 12o+48N @ 158o+1.15N @ 93o.=
Force of the squirrels.

X=63*cos12+43*cos158+1.15*cos93=21.69 N
Y=63*sin12+43*sin158+1.15*sin93=30.35 N.

tanA = Y/X = 30.35/21.69 = 1.39490
A = 54.45o.

Fs = X/cosA = 21.69/cos54.45=37.3 N. @ 54.5o = Force of the squirrels.

The force of the chipmunk should be equal and opposite from squirrels:

Fc=37.3N @ (54.5+180) = 37.3N.@ 234.5o
= Force of chipmunk.