Velocity Ratio, mechanical advantage, Efficiency of a screw jack

  • #1

Homework Statement:

Velocity Ratio, mechanical advantage, Efficiency in screw jack

Relevant Equations:

$$VR = \frac{2\cdot\pi \cdot R}{P}$$
$$M.A = \frac{/text{Load Lifted}}{\text {Effort Applied}}$$
$$\text{Efficiency} = \frac {M.A}{V.R}$$
Hello,

I have completed the question below.

I am just unsure on whether i am correct or not.

I am unsure on Mechanical Advantage. As i have seen a few different equations.

Question:

Screw Jack.png


Answer:

a)

Velocity Ratio = $$\frac{\text{Distance Moved By Effort}}{\text{Distance Moved By Load}}$$

$$VR = \frac{2\cdot\pi \cdot R}{P}$$

$$ R = 180mm$$
$$P = 12mm$$

$$VR = \frac{2 \cdot \pi \cdot180mm}{12mm}$$
$$VR = 94.2$$

b)

$$M.A = \frac{\text{Load Lifted}}{\text {Effort Applied}}$$

$$\text{200kg to Newtons} = 1961.33$$

$$M.A = \frac{ 1961.33}{80}$$
$$ = 24.51$$

c)

$$\text{Efficiency} = \frac {M.A}{V.R}$$

$$\frac{24.51}{94.2}$$

$$ = 0.26$$

Are these correct? If not where have i gone wrong?

Thank you.
 

Answers and Replies

  • #2
690
406
Good to see that you keep going, Ben. :smile:
Rather than the pitch, the problem shows only "The lead screw is 12 mm".
I guess we can assume that 12 mm is the pitch or gained height for each turn of the nut.

Your results are correct.
Why are you unsure on Mechanical Advantage?

This problem is asking about the actual mechanical advantage, which is a ratio of forces or loads.
The ideal MA is a geometric ratio of distances, arcs or velocities.

Copied from
https://en.wikipedia.org/wiki/Mechanical_advantage

"The ideal mechanical advantage (IMA), or theoretical mechanical advantage, is the mechanical advantage of a device with the assumption that its components do not flex, there is no friction, and there is no wear. It is calculated using the physical dimensions of the device and defines the maximum performance the device can achieve."
...
"The actual mechanical advantage (AMA) is the mechanical advantage determined by physical measurement of the input and output forces. Actual mechanical advantage takes into account energy loss due to deflection, friction, and wear.
The AMA of a machine is calculated as the ratio of the measured force output to the measured force input."

Please, note that this jack is a compound machine or a combination of two simple machines: a lever and a screw.

Copied from
https://en.wikipedia.org/wiki/Simple_machine

"A simple machine uses a single applied force to do work against a single load force. Ignoring friction losses, the work done on the load is equal to the work done by the applied force. The machine can increase the amount of the output force, at the cost of a proportional decrease in the distance moved by the load. The ratio of the output to the applied force is called the mechanical advantage."
...
"A compound machine is a machine formed from a set of simple machines connected in series with the output force of one providing the input force to the next."
...
"So in non-ideal machines, the mechanical advantage is always less than the velocity ratio by the product with the efficiency η. So a machine that includes friction will not be able to move as large a load as a corresponding ideal machine using the same input force."



Copied from
https://en.wikipedia.org/wiki/Screw_(simple_machine)

"It can be seen that the mechanical advantage of a screw depends on its lead,
l\,
. The smaller the distance between its threads, the larger the mechanical advantage, and the larger the force the screw can exert for a given applied force. However most actual screws have large amounts of friction and their mechanical advantage is less than given by the above equation."
...
"Because of the large area of sliding contact between the moving and stationary threads, screws typically have large frictional energy losses. Even well-lubricated jack screws have efficiencies of only 15% - 20%, the rest of the work applied in turning them is lost to friction. When friction is included, the mechanical advantage is no longer equal to the distance ratio but also depends on the screw's efficiency. From conservation of energy, the work Win done on the screw by the input force turning it is equal to the sum of the work done moving the load Wout, and the work dissipated as heat by friction Wfric in the screw."

For this screw jack, the input work (90 Newtons-meter or Joules) is the work done on the machine as the input force (80 Newtons) acts through the input distance (1.13 meters).
Nevertheless, the output work (23 Joules) is the work done by the machine as the output force (1962 Newtons) acts through the output distance (0.012 meters).
Hence, the efficiency of 26% that you have calculated (67 Joules of work will always be wasted, due to friction mainly, while lifting a mass of 200 Kg).
 
Last edited:

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