Question

A 5000 - ton ship rests on launching ways that slope down to the water at an angle of 10 ° . If the coefficient of sliding friction is 0.18 , how much force is needed to winch the ship down the ways into the water ?​

Answer 1

See below

Step-by-step explanation:

5000 ton = 10 000 000 pounds = 4,535,924 kg

Normal force = mg cos 10

then friction force = .18 * mg cos 10

the force downplane = mg sin 10

the force of the winch + force downplane must be = f of friction

w+ mg sin 10 = .18 mg cos 10

w = mg ( .18 cos 10 - sin 10 ) Plug in the numbers to find

w = 160 957 N

( this converts to 36185 pound-force)