Question

2. A sailor pushes a 100.0 kg crate up a ramp that is 3.00 m high and 5.00 m long onto the deck of a ship. He exerts a 650.0 N force parallel to the ramp. What is the mechanical advantage of the ramp? What is the efficiency of the ramp? Your response should include all of your work and a free-body diagram. Please help. I've been looking through notes and nothing makes sense.

Answer 1

The free body diagram makes a lot a sense to me but what im having a hard time with is the equation part. like putting the given answers into a equation to figure out the overall answer.

Answer 2

This isn't a free body diagram. And what equation are we talking about here?

Answer 3

like putting the number like the force and distance into a mechanical advantage equation and efficiency equation

Answer 4

Can you write those equations? I think I learned those but don't recall them at the moment

Answer 5

there's two theres real and ideal and the real mechanical advantage is MA= Fr/Fe and the ideal mechanical advantage is IMA= de/dr. efficiency is eff= MA/IMA x 100%.

Answer 6

Are you studying Work and kinetic energy?

Answer 7

I am studying work. I have online school it's called k12 and we do units and im on the second unit and its just about work power and mechanical advantages for like simple machines. I studied kinetic energy last semester.

Answer 8

I don't know about mechanical advantages and efficiency but I do know that regardless of the ramp, the person must do the same amount of work. Less FORCE is required with a longer ramp but the work is the same either way.

Answer 9

how would you figure out the less force? I've seen people try solving it and some how they come up with another force of 9.8 and I don't see how.

Answer 10

\[\huge W_{external} = W_{man} + W_{gravity}\]
There is no external work on the system
\[\huge W_{external} = W_{man} + W_{gravity}=0\]
\[\large W_{man} =- W_{gravity}= - (mg)L[cos(\theta +90)]
=mgLsin\theta = mgh\]

Answer 11

oh whoops, the last equation got split

Answer 12

\[\large W_{man} =- W_{gravity}= - (mg)L[cos(\theta +90)] =mgLsin\theta = mgh\]

Answer 13

so the gravity is 9.8 the L is 5.00 right?

Answer 14

yes

Answer 15

okay and to find the cosine you pretty much treat it like a right triangle and do 90 45 45?

Answer 16

umm you don't know the angles; if you do you didn't post it

Answer 17

I don't know them but I think I understand the equation but what would the mg be? would it be the 100 kg but converted into mg?

Answer 18

m=100kg
g=9.8m/s^2
L=5m
h=5 sin (theta)
theta=???
I'm thinking that since we know the length of the ramp and the height
\[\theta = sin^{-1}(5/3)\]

Answer 19

oh whoops, flip those

Answer 20

so \[\theta= 36.86989765 \] and if you round it, it would equal 36.87

Answer 21

hopefully you have answers to verify that

Answer 22

What do you mean?

Answer 23

Well, I don't want to claim that the answer are numerically correct just to find out they're wrong. I'm not that egotistical.

Answer 24

=)

Answer 25

lol okay well thank you very much for helping me understand this better I really appreciate it

Answer 26

uh huh