Can someone give me some pointers on this question...
Discussion
Hi all,
I'm having a little trouble with my college work at the moment, and was wondering if anyone can give me an equation, or series of equations to get the answer to this question:
A screw jack is 47.5% efficient, and has a thread with a pitch of 3.5mm. The radius of the tommy bar is 350mm. If a force of 120N applied to the tommy bar is necessary to raise a vehicle, what is the mass of the vehicle?
While this may sound silly, I'm really struggling to get my head round it! I'm not bothered about getting answers to the question, more the equations associated with being able to get the answer so I can try and understand it.
Thanks in advance.
I'm having a little trouble with my college work at the moment, and was wondering if anyone can give me an equation, or series of equations to get the answer to this question:
A screw jack is 47.5% efficient, and has a thread with a pitch of 3.5mm. The radius of the tommy bar is 350mm. If a force of 120N applied to the tommy bar is necessary to raise a vehicle, what is the mass of the vehicle?
While this may sound silly, I'm really struggling to get my head round it! I'm not bothered about getting answers to the question, more the equations associated with being able to get the answer so I can try and understand it.
Thanks in advance.
This is a glorified lever - its job in life is to increase the applied force by making you move it through a larger distance. So we just care about the ratio of the distance the applied force (eg your hand) moves to the distance the car moves.
Suppose you rotate the tommy bar once. The screw (and car) will advance by one thread, ie 3.5 mm. Your hand will have moved through a circle with radius 350 mm. This is a distance of 2*Pi*r, or about 2200 mm. Now you know the ratio of the two distances involved, you can work out the ratio of the forces.
But of course the jack is not completely efficient, so not all this force is due to the vehicle...
Hope this pushes you in the right direction, post again if still mystified.
Suppose you rotate the tommy bar once. The screw (and car) will advance by one thread, ie 3.5 mm. Your hand will have moved through a circle with radius 350 mm. This is a distance of 2*Pi*r, or about 2200 mm. Now you know the ratio of the two distances involved, you can work out the ratio of the forces.
But of course the jack is not completely efficient, so not all this force is due to the vehicle...
Hope this pushes you in the right direction, post again if still mystified.
Caveat - This could all be bks:
Assuming you don't know the diameter of the thread?
Ratio of the displacement between the effort applied and the movement of the load = 2*pi*R/p (Radius of the bar/pitch of the thread)
Ratio of the forces = Fload/Feffort
Efficiency = Work done at load/Work done by effort = (Fload/Feffort)/(2*pi*R/p)
Mload = 3650kg?
Assuming you don't know the diameter of the thread?
Ratio of the displacement between the effort applied and the movement of the load = 2*pi*R/p (Radius of the bar/pitch of the thread)
Ratio of the forces = Fload/Feffort
Efficiency = Work done at load/Work done by effort = (Fload/Feffort)/(2*pi*R/p)
Mload = 3650kg?
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