Views : 22,651,604
Genre: Science & Technology
Date of upload: Mar 29, 2018 ^^
Rating : 4.915 (6,724/308,240 LTDR)
RYD date created : 2022-04-09T21:33:48.982689Z
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1:36
Mark: But I could lift 10lbs 50 times.
Uses a 12 pound weight
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Hey Mark! Nice Video.
I wanted to point out that, at 2:03, mechanical advantage doesn't necessarily equal number of pulleys in all pulley systems. This is because fixed pulleys (pulleys which don't move while the load does) have a mechanical advantage of 1, while movable pulleys (pulleys which do move while the load does) have a mechanical advantage of 2. This is why, in your case, the pulley system had a mechanical advantage of 4: the specific arrangement and positioning of 2 movable and 2 fixed pulleys allowed the system to have a mechanical advantage of 4. However, it is important to note that this doesn't happen to all arrangements of 4 pulleys, as if you were to line up 4 fixed pulleys, you would still get an mechanical advantage of 1, not 4. Thus, you cannot say that mechanical advantage equals number of pulleys in all pulley systems. Instead, you could "split" up the tension along the ropes of a pulley through a diagram. Here's the procedure: you first can imagine that you are holding the end of whatever pulley system you are trying to solve with such a force such that the system is in static equilibrium. Next, you can split the tension caused by the force of the load among the ropes of the pulley system. The key to remember is that each unbroken piece of rope will have uniform tension along it. Finally, you keep splitting the tension all the way until you reach the rope which you are holding. The mechanical advantage of the pulley system will simply be the the load force divided by the effort force, which is simply equal to the tension in that rope you are holding, since the system is in static equilibrium. THAT is the correct way to solve for the mechanical advantage of a pulley system (sorry if I explained it badly).
Additionally, I wanted to point out that, at 2:08, mechanical advantage of an inclined plane does not equal the ratio of the length to the height unless the inclined plane is just flat and not inclined, in which case I wouldn't really call it an "inclined" plane. But instead, mechanical advantage of an inclined plane is equal to the ratio of the hypotenuse to the height. This can be derived like this: the load force on the inclined plane is simply equal to its downward force caused by gravity, which is simple equal to f = ma = mg. We can split this force into its x- and y- components, which we can then deduce that the force of the load in the direction parallel to and going down the hypotenuse of the inclined plane is equal to mgsin(theta), in which theta is the angle of the elevation of the inclined plane. This means that, if we were to push the load up on the inclined plane with force mgsin(theta), then the system would achieve static equilibrium. In other words, the load would stay in place. Since we know that mechanical advantage is defined as the ratio of the output force to the input force, that would simply mean that the mechanical advantage of the inclined plane would be mg/mgsin(theta), as mg is the output force in this case and mgsin(theta) is the input force in this case. This simplifies to 1/sin(theta). This is simply equal to the ratio of the hypotenuse to the side opposite angle theta, which in this case is height. Thus, mechanical advantage of an inclined plane is equal to the ratio of the hypotenuse to the height.
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@danieljensen2626
4 years ago
When we did this in highschool our speed/acceleration contest was only over a meter, so I just used the mouse trap to catapult my car over the finish line.
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