Free Falling Body

The maximal force depends on the deformation (stiffness) of the 4 suspenders.

As I do not know what you intend to use as suspender type it is quite difficult to give you an answer.

In a general form you should compute the energy gained by the mass considering its full travel and the elastic deformation energy of suspensions. The equation based on them being equal gives the maximal force.

Most physics teachers would want to see an answer that involved the elasticity of the four suspenders. Does that help?

I supposed that it is a home work the "dear friends" are asked to solve thus my general answer which in fact gives all indications but does impose a personal work for the questioner.

This assumes a 100% inelastic collision. If the weight rebounds, not all it's energy is transferred.

The solution suggested under #6 is not correct since it does not consider the way energy is stored in a deformable system (your suspension). And computes the energy gained by the mass in the gravitational field only on part of its travel. The correct result is much higher. One of the errors consists in assuming the force as constant over the deformation travel. The other is to assume that the mass does totally stop when it contacts the tray. Both assumptions are false. In fact an information was not present in the question it is the tray own mass since after the contact this mass will also move and for a 5000kg load mass the tray mass will not be so small to be neglected.

Help understand this. First, the calculated force is an average force.

However, the load has a kinetic energy and it is a finite amount. So, only so much energy can be imparted into the system (tray and suspension).

What do you mean that the mass does not stop when it contacts the tray? The term d is supposed to be the amount of movement or deformation in the tray and its suspension. So the formula assumes that there is a distance traveled after the impact. Isn't that the movement?

Lastly, I understand that the tray's mass and the suspension that the tray hangs from has mass, so we don't know the performance of the tray's movement, but that is why I framed the response with the term d for the displacement after impact.

I guess we may not be able to model the tray and the suspension, but if we assume that it moves and we can put a number on the amount of movement (even if it is arbitrary), the formula I used should be able to calculate the average force, don't you agree? If not, why not?

Just the fraction of the micron above the tray, the formula would be F=M*a.

This is correct and always will be correct. You are introducing aspects that are not within the question by considering the deformation of tray, rods or anything else.

Force is therefor easily calculated by above formula and what happens to the weight of whatever mass after that initial impact is not part of the question.

My knowledge is nearly 30 years old now so I could be out of touch but perhaps still correct

Answer to #8&9.

To comment #9: Please consider the last sentence in the question : "i want to know force coming on those suspenders"

According to my modest opinion the interest is to know how big the load on the suspenders will be since for a correct design the maximal load is required. It seems to me that Case 491 and myself have not same interpretation of the text. If it would be only a language problem I would not comment but it seems to me to be a technical aspect independent of the language. The interest is not to know how big is the weight of the mass which can be easily computed by the product Mxg. But of course I can be wrong as usual I never assume to detain the truth.

To comment #8: I made a sketch to explain why and where I consider the errors to be, it is enclosed. As mentioned above the average force is not of any interest since as you will see the ratio maximal to average with the values given and assumed by you is 2.5! Following your approach the suspenders would have been strongly under dimensioned with a high risk of destruction and accidents.

The used model was not complete and did not consider the movement of the mass after the contact. I used the linear force variation assuming that the structure will stay in the elastic range and will be used not only once.

In fact if the tray mass is also considered which would be according to my opinion compulsory the "stop force" will be different but to respect comment #9 why should I involve so many aspects since they were not mentioned in the question.

By the way I used the spelling checker

Also, distribution of impact force between four suspenders will depend upon the symmetry of the impact point wit respect to suspenders

Agree with 6 no answer. Force = m( V- 0)/ time( Taken to stop after impact)

I think a tray made to be structurally stiff enough (in this case, infinite) to catch the weight but not deform would weight so much that the momentum of the dropped weight would be divided between the two masses and the speed would be reduced significantly as the two objects continued their journey downward. Perhaps the tray was already accelerating, 'catching' the weight with zero impact at precisely 400mm and then slowing the weight down.

Anyway, this is straight out of grade 10 physics. Have fun with the homework. Kudos for using the web but we'll feed you enough reality to see that it's best just to suck it up and learn the theory the old fashion way like we all have. Keep up the hard work and we'll see you in a couple years.

In reply to #11

You are of course correct. It is in real design analysis even compulsory to go further into details as:

- rotation after impact if it is not centered

- influences of the different stiffnesses of the 4 suspenders even if the impact is centered

- stiffnesses of tray and connections between suspenders and structure

- stiffness of the structure

It is I assume a school home work and I did react only because in the computation of the average force the further movement of the falling mass was not considered in the energy balance.

I consider that if we assume the question is a home work we should not make computations but give to the student what indications he needs in order to solve the problem. Those indications should be in any case correct according to the physics of the problem.

Reply to#13

There is no "perhaps", at impact the tray is accelerated and the falling mass decelerated and assuming as "soft" impact they will have same speed. This is the reason I mentioned that if the tray mass would be considered the maximal force will be different. Concerning your last sentence: what would have accelerated the tray to have same speed as falling mass and avoid an impact as they meet?

I agree with the #6 of Hero. and vote. He works out the force from work and energy relationship, we can also get from momentum law. means forcexacting time= changes of momentum.

Howver this is view from average value, if we take account of instant force, we can arrang movement equation group. and initial condition is the end speed of the object falling. as we know the smaller elastricity coefficence,the more deformation the spring (suppose the suspende is spring or something like flexible threads) is getting, the distance is longer. the object gets smaller reaction force .as matter of fact, the spring become a buffer.

we get the sin movement expression from the spring elasticity equation so tht the force is also changing as sin regularity.

the total elastricity coef. = 4 x unit suspender elas. coef.

there are serveral situation,

1, elas coef =< critical value, the ball moves according to the board,

2. it > c. v. the ball will rebound repeatly on the board.

3. the more complex situation will be expressed by specialists.