
#1
Aug1405, 11:27 PM

P: 28

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.




#2
Aug1405, 11:50 PM

P: 230

Are you sure about the question?




#3
Aug1505, 07:44 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

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", F_{e}, acts over distance x_{e}, applying force, F_{a}, over distance x_{a}, then by conservation of energy, we must have F_{e}x_{e}= F_{a}x_{a} so that [itex]F_a= (\frac{x_e}{x_a})F_e[/itex] : the (theoretical) mechanical advantage is [itex]\frac{x_e}{x_a}[/itex] 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"! 



#4
Aug1505, 09:15 PM

P: 28

mechanical advantageAn 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? 



#5
Aug1505, 09:45 PM

Emeritus
Sci Advisor
PF Gold
P: 2,977

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? 



#6
Aug1605, 07:17 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

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. 



#7
Aug1605, 07:28 PM

P: 28

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 



#8
Aug1605, 09:16 PM

Emeritus
Sci Advisor
PF Gold
P: 2,977





#9
Aug1705, 06:33 AM

P: 28

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




#10
Aug1705, 10:04 AM

P: 230

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. 



#11
Aug1705, 01:38 PM

Mentor
P: 40,907

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. 


Register to reply 
Related Discussions  
Mechanical Advantage of Pulley  Introductory Physics Homework  4  
Mechanical Advantage : Please Help!  Introductory Physics Homework  1  
Unit of Mechanical Advantage of a machine?  General Physics  2  
Mechanical Advantage  Introductory Physics Homework  2  
how to improve the mechanical advantage of a catapult  Classical Physics  2 