Physics - Direction of a barge?

K

Kiefer

New member
Hmm. Well, you want the x-components of these two forces to cancel, so that the barge moves only northward and not east or west. Then, the sum of the forces acting on the barge is going to be...

ΣFx= (80N)(cos60) - (120N)(cosx) = 0
(80)(cos60) = (120)(cosx)
cosx = (80)(cos60)/(120)
x = cos^'1 (80)(cos60)/(120)
x = 70.53 degrees...

As far as I can tell, the second cable should be pulled in a direction (90 - 70.53) = 19.47 degrees east of north. Where did you get your answer? Either I'm wrong, or it is.
 
A 1000kg barge is being towed by means of two horizontal cables. One cable is pulling with a force of 80.0N in a direction 30.0 degrees west of north. In what direction should the second cable pull so that the barge will accelerate northward, if the force exerted by the cable is 120N?

The answer is 39.0 degrees east of north. My professor showed me to do sin^-1(80/120*sin30) but that doesn't work. Any help here guys?
 
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