How Far Does a 3.5 Mg Engine Hoist in 4 Seconds?

[fball558]

Homework Statement

a 3.5 Mg engine is suspended from a spreader beam having a negligible mass and is hoisted by a crane which exerts a force of 40kN on the hoisting cable. determine the distance the engine is hoisted in 4 s starting from rest.

Homework Equations

f=ma
vf^2 = vi^2 + 2ad
v=Vo + at

The Attempt at a Solution

first i used f=ma
converted force of 40kN to N so 40000 and converted 3.5 Mg to kg so 3500
then a = f/m so a = 11.4286 m/s^2 (this seems to high but i continued anyways)

then i took v= Vo +at Vo=0 a is from above and t = 4s just plug in and you get 45.71 m/s

then use
vf^2 = vi^2 + 2ad vf = 45.71 vi = 0 a = 11.43 solve for d and you get something like 91m that i know is wrong. i should get something like 13 m does anyone see where I am going wrong??

 

[srmeier]

There is more than one force acting on the engine. You are only accounting for the force of the crane that is why the net force is high.

 

[fball558]

oh ok... i got it now i forgot to take away the force of the crate pulling down on the crane. when i do that i get the right answer now.
thanks a lot srmeier!

 

[srmeier]

My pleasure ^^

Side note: (One force is caused by the collection of hoisting cables which are connected to the engine & the other is the force of gravity on the engine. also note that the tension force is only in the y-direction because the x-components of force cancel one another.)