Posted by **Emilie** on Wednesday, February 23, 2011 at 11:07pm.

Two prospectors are pulling on ropes attached around the neck of a donkey that doesn’t want to move. One

prospector pulls with a force of 25 lbs, and the other pulls with a force of 50 pounds. If the angle between the

ropes is 30 degrees, then how much force must the donkey use in order to stay put? (The donkey knows the

proper direction in which to apply his force.)

- Linear Algebra -
**Anida**, Sunday, April 20, 2014 at 8:16pm
Assuming the forces are acting 12.5 degrees (25/2) out from the head of the donkey in either direction then the forward force would be 55 * cos12.5 + 75 * cos12.5 = 127 N to 3 s.f. so the donkey pulls backwards with 127 N - but this depends entirely on the orientation of the diagram of forces, if there is one

- Linear Algebra -
** Henry**, Tuesday, December 22, 2015 at 10:37pm
Since the system is in Equilibrium, the

force of the donkey must be equal and opposite the applied force:

Fd = -(25 + 50[30o]) = -(25 + 43.3+25i) = -(68.3 + 25i) = -68.3 - 25i = 72.7 Lbs[20.1o] S. of W.

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