25

From the "Solo Traveler" patent: A more specific object of the present invention is to provide a lid having an opening formed therethrough to enable drinking, and having a recess formed in the lid adjacent the opening to accommodate the upper lip of one drinking from the cup This is also illustrated in the patent diagrams. I came across this interesting ...


18

Once the cylinder is released and starts to accelerate downward, the rotating flyweights should pivot upwards to a certain degree (as shown in the drawing) and should act as a drag on each of the motors' rotor shafts and thus a drag on the falling cylinder No, there is no reason for them to pivot upwards. The gravity pulls the flyweights with the same ...


13

radian is a derived unit, defined as the ratio of arc length to radius. As the ratio of two lengths it is dimensionless.


12

Probably the easiest practical method to measure the frequency of a spinning top with everyday equipment is to analyze the sound created by the motion and look for the characteristic frequency. This can be done with a spectrum analyzer app that should be available for free for most smartphones. Place the spinning top close to the microphone of the phone and ...


11

I think you're overcomplicating things. To push 85 kg up a 15% slope against gravity of 9.8 m/s2 requires a force of $$ \sin (\arctan (0.15) ) = 0.1483 \approx 0.15 $$ $$F = 85 kg \cdot 9.8 \frac{m}{s^2} \cdot 0.15 = 125 N$$ With an 80 mm wheel, this requires a torque of $$T = 125 N \cdot 0.04 m = 5 Nm$$ To do this at a forward speed of 11.11 m/s ...


11

Sorry for not taking your "Thin Hollow Brick-Shaped Object" as an example - I felt a standard notched test specimen would illustrate the principle more clearly: It can be seen that adding external radii to the test sample has no effect on the stress. In this example, cutting even more material away from the corners would make no difference to the failure of ...


10

First, your statement is not universally true. Motion of the bottom of a building can be greater or lesser in upper floors. It depends on the type of motion, how free the building is to move in other axes, where you are relative to the building center of mass, and whether the motion hits any resonant frequencies of the building. If the motion excites a ...


10

I've been doing large pressure vessel design for a while, and ultimately it comes down to what is negligible v. what is not. Especially in the realm of atmospheric storage vessels, scrubbers, or vessels within 10 atmospheres of pressure, there isn't a lot of radial stress compared to the other conservative factors. In the realistic engineering world, by ...


9

This is a constant acceleration problem on an inclined plane which ignores friction. The force acting on the truck is the force of gravity. The acceleration the truck will experience will be solely due to gravity, $9.8\text{ m/s}^2$. Using the 38 degrees for the slope you have to find the component of gravity/acceleration going down the slope. Given the ...


9

The torsion constant $J_T$ relates the angle of twist to applied torque via the equation: $$ \phi = \frac{TL}{J_T G} $$ where $T$ is the applied torque, $L$ is the length of the member, $G$ is modulus of elasticity in shear, and $J_T$ is the torsional constant. The polar moment of inertia on the other hand, is a measure of the resistance of a cross ...


9

That figure is about right for a low tensile bolt. See also this calculator and this table As a reality check if we approximate to a cross sectional area of 7 mm2 and a load of 1000 N that gives a tensile stress of 140 MPa which is below yield even for low tensile steels. In this particular context, where torque is known, the thread pitch doesn't come ...


9

You are correct. A channel section in a "u" or "n" orientation is obviously asymmetric around its transversal x-axis. This means that its centroid is not located at the midpoint of its height. Instead, the centroid will be closer to its web than to the opening. Since the stress at a given fiber at a distance $y$ from the centroid is equal to $$\sigma = \...


8

You could have active suspensions, which could keep the vehicle level or even roll it into the corner (see for example this PhD thesis), but they are typically prohibitively expensive and not feasible in practice. Semi-active suspensions, which can lock the anti-roll bar solid just ahead of a turn, are sometimes a good compromise between performance and ...


8

I spent some time searching on this one, and surprise surprise, Imperial units strike again! The good old "pounds-force pounds-mass issue" masked with ounces. This is why any calculation more involved than sizing a pulley should be done in metric, lol. Oriental Motor- Basics of motor control, explains the following: Units of Measure for Moment ...


8

The intuitive meaning of Saint Venant's principle is that "if two different sets of applied loads are statically equivalent, then the differences between the two stress patterns they create are only significant close to where the loads are applied". This is used very frequently in engineering modelling. For example if a beam is bolted to a support with ...


7

There are several ways. The most direct is to paint half of the top white and the other black, then point something with a light sensor at it. AC amplify the output a little, then determine the frequency. This could be with a deliberate frequency counter, a scope that displays frequency directly, or a microcontroller you program to find the period and ...


7

The problem here is that $dr$ is not the radius, $r$ is! Your expression for the infinitesimal area should be $2\pi r dr$: this is for a thin loop of radius $r$. (Excuse the crude diagram): So $r$ is a coordinate which is the radius of the loop, and $dr$ is a infinitesimal increment of $r$, i.e. the thickness of the loop. Since the loop is infinitely thin , ...


7

To add to @alephzero's answer, here's an layman's explanation of Saint-Venant's Principle: far enough away from the load's point of application, equivalent loads can be treated as identical. For instance, the stress profile of a column of area $A$ with a concentrated load $F$ applied at the top is effectively identical to the stress profile of the same ...


6

Harmonic coupling with the building's natural vibration frequencies, dissipation of energy in a massive building, and the variety of vibration modes that can experience coupling could result in more or less displacement and more or less velocity at lower floors or upper floors. There's no universal relationship between either velocity or displacement ...


6

You need to control how fast you apply the power. Some sort of damper mechanism that would allow power to be gradually applied and not cause it to spin out. You could try and wrap the 'power string' over different sized pulleys so that you changed the gear ratio. If you used a cone it would be similar to a CVT where starting out you would have a low gear ...


6

That model of encoder is not intended for mechanical measurements. It's meant to be used as a fancy knob with high friction, for turning by hand, say, to pick a position from menu on a display, and trying to couple it to a mechanism for readout is a misguided attempt. What you need is an encoder based on slotted optocouplers or a reflective optocouplers. ...


6

Fred's answer solves this problem using classical mechanics/kinematics. However, you can also solve this type of problem using an energy approach (which I have always found easier to do - Newton's laws just feel more intuitive). Newton's first law (conservation of energy) The baseline potential energy (zero) is the truck's starting position, and the final ...


6

No. Your basic problem is that "torque is proportional to radius" is wrong. Torque is the expression of normalized rotational force independent of radius. For example if a motor can deliver 10 Nm torque, then a pulley with 1 m radius can produce 10 N of force, at 500 mm radius 20 N, at 2 m radius 5 N, etc.


6

Gear ratios can be referred to as hunting and non-hunting ratios. In a non-hunting ratio, any one gear tooth will contact its corresponding gears' teeth in the same place every time. In a hunting ratio, the number of teeth in the gear set have no common factors between them. This way as the pitch diameters 'roll' over one another it takes a very long time ...


6

If we're staying within the realms of beam theory, we can go with this approach which is valid for any material that exhibits linear-elastic behaviour before yield: The curvature of a beam is related to the applied moment and it's flexural stiffness: $$ \begin{align} \kappa = \frac{M}{EI} \end{align} $$ where, for a cicular rod, $I = \frac{\pi d^4}{64}$. ...


6

jpa made a good answer on why the weights wouldn't flex upwards. I'll add another point. Newton's third law says that for every action, there is an equal and opposite reaction. For the box to be slowed by this mechanism, it must exert a force on something else, in the same way gravity pulls up on earth and aerodynamic drag moves the air. However, this ...


5

The yield point of a material is determined by looking at its stress-strain graph. As you can see from the image, some materials do not exhibit an obvious point where the material starts yielding. When this is the case, the "yield point" is determined by the 0.2% offset rule. This is shown by the dashed line in the graph. As you can see from the graph, even ...


5

One way to think of MEMS is regular mechanical systems, but at a very small scale. Often these tiny systems are fabricated using technology developed for making silicon electronic chips, like nanometer-scale photolithography and etching. However, describing MEMS as just downscaled regular mechanical systems is doing the concept injustice. Various physical ...


5

You can't calculate friction coefficients this way. At the atomic-scale level, the friction coefficient depends on the interactions of the two materials across the interface. If you measure the behaviour of the interface between materials A and B, and A and C, you can't say anything about the interface between B and C. In any case, friction coefficients are ...


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