Could someone help me with this, please. Thanks.
The tension in a rope attached to a boat is 27.5 lb. The
rope is attached to the boat 6.0 ft below the level at
which the boat is being drawn in. At the point where
there is 18.0 ft of rope out, what force is bringing the
boat towards the dock, and what force tends to raise
the boat? What if there is 9 ft of rope out?
2 answers
confused--Please help!!!!
With 18 feet out, the angle the rope makes with the horizontal is
sin^-1(6/18) = sin^-1 (1/3)
= 19.5 degrees
Horizontal component of rope tension =
27.5 cos19.5 = 25.9 lb
Vertical component of rope tension =
27.5 sin19.5 = 9.2 lb
Proceed similarly for the 9 ft rope case.
sin^-1(6/18) = sin^-1 (1/3)
= 19.5 degrees
Horizontal component of rope tension =
27.5 cos19.5 = 25.9 lb
Vertical component of rope tension =
27.5 sin19.5 = 9.2 lb
Proceed similarly for the 9 ft rope case.
3 answers