Let's kick things off with some fairly easy questions, just to get the ball rolling.
1) Explain briefly why an airplane with low-mounted engines pitches up with the throttle is increased while an airplane with high-mounted engines may pitch down when the throttle is increased.
2) What does the lowercase Greek letter Eta (η) stand for, in the context of this event?
3) If I weigh 100 kg and the object I'm trying to move weighs 5.0E4 kg, and the Class 1 lever is 20 m long and massless. Myself and the object are at opposing ends of the lever. How far from the object, in meters, must the fulcrum be such that when I stand on the opposing end of the lever, we are in static equilibrium? (If it is impossible, note that)
a) If the lever is changed to Class 2, does this number change? Assume the forces are now acting in the correct direction to achieve equilibrium.
b) Ditto, but with Class 3.
4) What is the difference between Pitch and Lead? (Screws, for Division C)
Assassinator wrote: ↑September 9th, 2020, 6:52 am
Let's kick things off with some fairly easy questions, just to get the ball rolling.
1) Explain briefly why an airplane with low-mounted engines pitches up with the throttle is increased while an airplane with high-mounted engines may pitch down when the throttle is increased.
2) What does the lowercase Greek letter Eta (η) stand for, in the context of this event?
3) If I weigh 100 kg and the object I'm trying to move weighs 5.0E4 kg, and the Class 1 lever is 20 m long and massless. Myself and the object are at opposing ends of the lever. How far from the object, in meters, must the fulcrum be such that when I stand on the opposing end of the lever, we are in static equilibrium? (If it is impossible, note that)
a) If the lever is changed to Class 2, does this number change? Assume the forces are now acting in the correct direction to achieve equilibrium.
b) Ditto, but with Class 3.
4) What is the difference between Pitch and Lead? (Screws, for Division C)
1. No clue, Google isn't turning up anything either. I'll leave this one to the next person.
2. Efficiency, the ratio of actual mechanical advantage to ideal.
3. 50000x = 100(20-x)
500x=20-x
501x=20
x=.0399
.040 m
3. b. I assume the rock is on top, providing (5e4)g newtons of force downward, but... how much force upward are you exerting? Same goes for c.
4. Pitch is the distance between adjacent threads, as seen from the side of the screw. Lead is the distance downward the screw moves for every rotation. This is only different on a multiple-start screw.
Re: Machines B/C
Posted: September 9th, 2020, 9:50 am
by CPScienceDude
I think for #1. . .
I'm thinking for 1 it's because they're not on the axis of the center of mass so their thrust isn't evenly applied. I worded that weird but yeah
Re: Machines B/C
Posted: September 9th, 2020, 10:10 am
by Creationist127
CPScienceDude wrote: ↑September 9th, 2020, 9:50 am
I think for #1. . .
I'm thinking for 1 it's because they're not on the axis of the center of mass so their thrust isn't evenly applied. I worded that weird but yeah
That makes sense--the thrust becomes more of a torque than a linear force. I was going at it with the assumption that the engine was the center of mass... which really isn't true.
Assassinator wrote: ↑September 9th, 2020, 6:52 am
Let's kick things off with some fairly easy questions, just to get the ball rolling.
1) Explain briefly why an airplane with low-mounted engines pitches up with the throttle is increased while an airplane with high-mounted engines may pitch down when the throttle is increased.
2) What does the lowercase Greek letter Eta (η) stand for, in the context of this event?
3) If I weigh 100 kg and the object I'm trying to move weighs 5.0E4 kg, and the Class 1 lever is 20 m long and massless. Myself and the object are at opposing ends of the lever. How far from the object, in meters, must the fulcrum be such that when I stand on the opposing end of the lever, we are in static equilibrium? (If it is impossible, note that)
a) If the lever is changed to Class 2, does this number change? Assume the forces are now acting in the correct direction to achieve equilibrium.
b) Ditto, but with Class 3.
4) What is the difference between Pitch and Lead? (Screws, for Division C)
1. No clue, Google isn't turning up anything either. I'll leave this one to the next person.
2. Efficiency, the ratio of actual mechanical advantage to ideal.
3. 50000x = 100(20-x)
500x=20-x
501x=20
x=.0399
.040 m
3. b. I assume the rock is on top, providing (5e4)g newtons of force downward, but... how much force upward are you exerting? Same goes for c.
4. Pitch is the distance between adjacent threads, as seen from the side of the screw. Lead is the distance downward the screw moves for every rotation. This is only different on a multiple-start screw.
CPScienceDude wrote: ↑September 9th, 2020, 9:50 am
I think for #1. . .
I'm thinking for 1 it's because they're not on the axis of the center of mass so their thrust isn't evenly applied. I worded that weird but yeah
1. Yep.
2. Yep.
3. Yep. (Unless my answer 'key' is also wrong)
a) and b) Assume that the effort is equivalent to what 100 kg of "inverted gravity" would be. I also only now realized for the Class 3 lever, you will be solving the distance from the fulcrum of the effort, not the load, as the load will be placed at the other end of the lever. Whoops.
4. Yep.
Assassinator wrote: ↑September 9th, 2020, 6:52 am
Let's kick things off with some fairly easy questions, just to get the ball rolling.
1) Explain briefly why an airplane with low-mounted engines pitches up with the throttle is increased while an airplane with high-mounted engines may pitch down when the throttle is increased.
2) What does the lowercase Greek letter Eta (η) stand for, in the context of this event?
3) If I weigh 100 kg and the object I'm trying to move weighs 5.0E4 kg, and the Class 1 lever is 20 m long and massless. Myself and the object are at opposing ends of the lever. How far from the object, in meters, must the fulcrum be such that when I stand on the opposing end of the lever, we are in static equilibrium? (If it is impossible, note that)
a) If the lever is changed to Class 2, does this number change? Assume the forces are now acting in the correct direction to achieve equilibrium.
b) Ditto, but with Class 3.
4) What is the difference between Pitch and Lead? (Screws, for Division C)
1. No clue, Google isn't turning up anything either. I'll leave this one to the next person.
2. Efficiency, the ratio of actual mechanical advantage to ideal.
3. 50000x = 100(20-x)
500x=20-x
501x=20
x=.0399
.040 m
3. b. I assume the rock is on top, providing (5e4)g newtons of force downward, but... how much force upward are you exerting? Same goes for c.
4. Pitch is the distance between adjacent threads, as seen from the side of the screw. Lead is the distance downward the screw moves for every rotation. This is only different on a multiple-start screw.
CPScienceDude wrote: ↑September 9th, 2020, 9:50 am
I think for #1. . .
I'm thinking for 1 it's because they're not on the axis of the center of mass so their thrust isn't evenly applied. I worded that weird but yeah
1. Yep.
2. Yep.
3. Yep. (Unless my answer 'key' is also wrong)
a) and b) Assume that the effort is equivalent to what 100 kg of "inverted gravity" would be. I also only now realized for the Class 3 lever, you will be solving the distance from the fulcrum of the effort, not the load, as the load will be placed at the other end of the lever. Whoops.
4. Yep.
1. a. 100g*20 = 50000g*x
20 = 500x
.04 = x
.040 m
b. 50000g*20 = 100gx
10000 = x
It is not possible to be in equilibrium.
1. No clue, Google isn't turning up anything either. I'll leave this one to the next person.
2. Efficiency, the ratio of actual mechanical advantage to ideal.
3. 50000x = 100(20-x)
500x=20-x
501x=20
x=.0399
.040 m
3. b. I assume the rock is on top, providing (5e4)g newtons of force downward, but... how much force upward are you exerting? Same goes for c.
4. Pitch is the distance between adjacent threads, as seen from the side of the screw. Lead is the distance downward the screw moves for every rotation. This is only different on a multiple-start screw.
CPScienceDude wrote: ↑September 9th, 2020, 9:50 am
I think for #1. . .
I'm thinking for 1 it's because they're not on the axis of the center of mass so their thrust isn't evenly applied. I worded that weird but yeah
1. Yep.
2. Yep.
3. Yep. (Unless my answer 'key' is also wrong)
a) and b) Assume that the effort is equivalent to what 100 kg of "inverted gravity" would be. I also only now realized for the Class 3 lever, you will be solving the distance from the fulcrum of the effort, not the load, as the load will be placed at the other end of the lever. Whoops.
4. Yep.
1. a. 100g*20 = 50000g*x
20 = 500x
.04 = x
.040 m
b. 50000g*20 = 100gx
10000 = x
It is not possible to be in equilibrium.
ID clue: sassy does Machines.
Good. Your turn!
Re: Machines B/C
Posted: September 17th, 2020, 9:17 pm
by RiverWalker88
Revive!
Two ideal, frictionless doorstops are arranged in such a way the one doorstop (Doorstop A) is resting on top of the other (Doorstop B), as shown below. The corner of Doorstop A is fixed on the floor so that the doorstop can rotate about that point, but cannot slide. A mass (m) is fixed on top of Doorstop A, and it has a mass of 0.55kg.
Doorstop A is made of Styrofoam and has a negligible mass, but Doorstop B is made of solid iron and has a mass of 1.17kg.
In this configuration, with Doorstop A resting on Doorstop B, what is the acceleration of Doorstop B?
After Doorstop B slides out from underneath Doorstop A, the end of Doorstop A that was resting on Doorstop B strikes the floor. This breaks the glue that held the mass to the doorstop, and the mass begins to slide down the doorstop. How long will it take the mass from its fixed position to slide all the way down Doorstop A such that the center of the block is no longer on the ramp? Assume that the block does not change its acceleration the entire time.
Re: Machines B/C
Posted: September 18th, 2020, 4:57 am
by astronomybuff
Looks very interesting. I'm really new to machines, but I'll give it a shot.
a. We can take the fixed point as the axis of rotation, so the torque is 0.05mg. Now, I'm a bit confused if this torque is what moves Doorstop B, but I'll just take a guess. a=F/m=0.05mg/M= 4.6 m/s^2.
b. Uh I hope the acceleration is just mg cuz there's no angle. x-x0=0.5at^2, .2=at^2,t^2=0.037, t = 0.19 seconds. Yeah... that doesn't seem very realistic .
Re: Machines B/C
Posted: September 18th, 2020, 8:45 am
by RiverWalker88
Good attempt! This problem was pretty involved (and needed some trig).
Part A:
I think this is how this problem is solved. My logic might be incorrect.
Part B:
I might have made a mistake or incorrect logic or something on these. I'm not 100% certain.
Your Turn!
Edit: Yes, I was too lazy to try to transcribe my work into an explanation. I apologize for the enormous images.
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