# Tag Info

6

Does anyone know if this would be a good idea? The proposed design has poor rigidity towards torsional loads. It doesn't matter whether the actuator is off-center or if the load is off-center: if load's center of gravity and actuator are not at the same position, the torsional load will likely damage the bearings. In your example with two guide rails, ...

6

Your mechanism is going to lock up like the old car jacks immediately, due to differential raising of the platform. I am sorry but it looks like a textbook example of what not to do. in most real-life cases where the lifting actuator has to be off-center, they use a linked pair of scissors mechanism, like what they do in trucks' gate lifter, or the hydraulic ...

5

In a scissors jack design, items one and three, you'll have the least mechanical advantage when the assembly is retracted. Using drawer slides or the equivalent is a good idea, but will not provide for a balanced lift if there are loads applied that are not uniform over the cabinet's bottom surface. Your selected motor, with a thirty to one reduction should ...

2

Obviously cranks and pistons have been used for hundreds of years for this. Your other option would be fast-acting motors or hydraulic rams. But a 500N (or is it 600?) force isn't what happens when something hits an object. Hitting an object imparts kinetic energy, and a 1 HP motor just doesn't have a lot to give because you have to speed something up so ...

2

Initial thoughts? I'm not a mechanical engineer but I think you'd be disappointed. Figure 1. A quick pictorial analysis shows that there is almost no reason for the lift arm to move up. Also, most of the effort in the linear actuator would be taken up by a force vector from C to A with very little of the effort going into rotation. Figure 2. An online ...

2

The idea is mostly sound. I have a load F that is too high for my linear rails....so instead of beefing up my rails, I increase the load sharing across 3 or more rails, thus dividing it. In reality, I have actually seen such things....but only on poorly built machines as a half solution. In general one can do anything they want....but that doesn't mean ...

2

IMHO (and I guess you already know that), from a structural point of view putting 100 kg at 1/4[m] will create a huge amount of bending moment on a actuator that it is rated for 120[kg]. An although you don't clarify whether the actuator electric/hydraulic, if bearings are involved you are bound to get failure sooner rather than later. Normally this type of ...

2

Either one would probably work as long as you gear it properly; but like Eric Shain, I'd advise you to go with the scissor jack because it's simpler. However, because the force varies with the angle of the scissors, you'll need to gear it down somewhat more than my suggestion. Lifting the setup is around 201 joules (41 kg * 0.5 m * 9.8 m/s (g)), and it would ...

2

Take a look at the illustration below, which is how I understand your question. You're asking, what would be the minimum value for the blue arrows in order to make each object move? First, let's look at the simpler case - the red box. I don't think this needs another diagram - you've rightly stated in your question that if there is a coefficient of friction ...

2

As a quick basic estimation approach we annotate the following: The length of bar 3 = L Top and bottom bearing force on each box $F_b$ The height and width of boxes 4 and its counterpart on channel 2 H and W. $$\Sigma M=0 ,\quad 220*L/4 = H/2*4F_b \rightarrow \quad F_b=220*L/8H$$ Now you have to figure if the bearings and their connections are okay to ...

1

What happens in the arrangement is that you are applying a force away from the center of gravity. Imagine you have a pencil on a flat surface and you try to push. If you don't push it around the center of gravity you will cause it to rotate. Although using linear bearings (especially longer ones) will mitigate the problem, you will be creating more normal ...

1

I might look into medical equipment, I was looking at actuators the other day, and they weren't too expensive.

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This is one of Newton's Axioms. I.e., strictly speaking the rule does not need to be able to be derived from some other physical law, but an empirical fact supported by a huge number of actual and possible experiments. This axiom (together with the second of Newton's Axioms), leads to the inertial principle - the total momentum of a closed system is ...

1

Rather than think of it as 'resisting', think of it as force. In a vacuum, you are correct that there is nothing to push against, nothing resisting. The resisting, as you put it, comes from pushing out something that was taken into space with the rocket, the gas from the rocket motor. In force terms, it takes force to push that gas out. The force is in ...

1

Let's call the point where the hinge is welded to the boom D. Scaling CD from your diagram we pick 30mm. That means the mechanical leverage of this gadget is approximately $$F_{boom}= F_{actuator}\frac{30}{400}= 0.075F{actuator}$$ This gets even worse because we need to consider the reduction of leverage due to the location of D, a nearly ratio of 1/20 ...

1

Use a screw press mechanism and reducing gearbox. Then it becomes an issue of calculating the force multiplication of the screw and gearbox mechanism to get 500 newtons. It won't be fast, but it will provide the force you require.

1

If budget was not an issue you could design a crank / piston assembly from scratch... As it is an issue then consider re-purposing a small single cylinder engine - lawnmower or chainsaw perhaps. Consider lubrication as well...

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