# Mechanical advantage

An effort of 3000kg is required to move a mass of 2000kg in a certain simple machine. If the mass is raised 1.5m while the effort moves 12m, find the actual mechanical advantage.

## Answers and Replies

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HallsofIvy
Science Advisor
Homework Helper
An effort of 3000 kg? kg is a unit of mass, not force. Perhaps you mean the weight of a 3000 kg mass (3000g Newtons) at sea level. In any case, it may not be relevant. The "mechanical advantage" is the force actually applied divided by the force exerted. If the "effort", Fe, acts over distance xe, applying force, Fa, over distance xa, then by conservation of energy, we must have Fexe= Faxa so that $F_a= (\frac{x_e}{x_a})F_e$ : the (theoretical) mechanical advantage is $\frac{x_e}{x_a}$ whatever the forces are.

In this particular case, that would be [itex]\frac{12}{1.5}[/tex] which clearly doesn't give 2000g Newtons as the result of an effort of 3000g Newtons.

Using the fundamental definition of "mechanical advantage": force applied divided by effort, we would get 2000g/3000g= 2/3 as the "actual" mechanical effort. Must be one heck of lot of friction in that system: although the theoretical mechanical advantage is 12/1.5= 8, the actual mechanical advantage is 2/3! Not much of an "advantage"!

quark said:
Are you sure about the question?
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An effort of 3kN is required to move a mass of 2000kg in a certain simple machine. If the mass is raised 1.5m while the effort moves 12m find the actual mechanical advantage.

The book gives an answer of 6.54! Do you have an explanation for the book's answer?

SpaceTiger
Staff Emeritus
Science Advisor
Gold Member
scientist said:
The book gives an answer of 6.54! Do you have an explanation for the book's answer?

mechanical advantage = load/effort

Both the "load" and the "effort" are forces. You're given the effort, but you have to calculate the load from the mass of the object. The load here can be thought of as the force required to lift the object if the machine weren't present. Any idea on how to get a calculate the load force from the given mass?

HallsofIvy
Science Advisor
Homework Helper
This is not at all what you originally posted!

However, the load divided by the effort is indeed 6.54! As SpaceTiger said, convert the load to Newtons. Once again, the distances given are irrelevant.

HallsofIvy said:
This is not at all what you originally posted!

However, the load divided by the effort is indeed 6.54! As SpaceTiger said, convert the load to Newtons. Once again, the distances given are irrelevant.
-------------------------------------------------------------------------

Space Tiger, can you check my answer?

load
MA= -----
effort

2000kg x 9.81N
= --------------
3000kg = 3kN

= 19620N
-------
3000kg

= 6.54
MA = 6.54

Is this correct? Please post with a reply.
Scientist

SpaceTiger
Staff Emeritus
Science Advisor
Gold Member
scientist said:
2000kg x 9.81N
= --------------
3000kg = 3kN

Your answer is right, but you seem to be confusing the definition of kg and kN. They're not the same thing. In fact, kg is a unit of mass and kN is a unit of force. They both have "k", which is short for "kilo", which is itself a prefix to indicate 1000. Since "g" is short for grams and "N" is short for Newtons, the kg and kN are 1000 grams and 1000 Newtons, respectively.

Ok space tiger, I can differentiate between the kg and kN. Is my method correct?

Scientist,

Follow SpaceTiger's advice and you will end up with correct procedure.

2000kgx9.81m/s^2 = 2000x9.81 N. (this is the right way of writing an equation if you want to score full points)

PS:Just keep in mind the dimension equality and you will never have troubles.

BTW, mechanical advantage can also be calculated by distances, theoretically. Both should give you same result theoretically. But in real life, resistance and friction come into picture and that is why we base it on forces.

Doc Al
Mentor
quark said:
BTW, mechanical advantage can also be calculated by distances, theoretically. Both should give you same result theoretically. But in real life, resistance and friction come into picture and that is why we base it on forces.
As SpaceTiger stated, mechanical advantage is just load/effort. Distances don't enter into it. If the actual effort force is used in the calculation, friction effects are included.

However, distance does enter into a calculation of efficiency, defined as "work out"/"work in". An ideal (frictionless) machine would have an efficiency of 1 (or 100%); real machines always have an efficiency less than 1.