User Name Remember Me? Password

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Apr 1st 2018, 11:34 AM #1 Junior Member   Join Date: Mar 2018 Posts: 12 Required force to move the object Hi Woody, I want to calculate the required force to move the object (70 Kg) for 1 sec at 22.22 m/s2. What is the distance traveled by the object for 1 sec at the same acceleration? 1. A Vehicle running at uniform velocity of 80 KMPH (50 MPH) i.e 22.22 m/s. 2. No acceleration since uniform velocity. 3. It picks the object while running. 4. What would be the acceleration of the Object? = 22.22 m/s2 ? as per the calculation that I mentioned above. 5. The object should accelerate at 22.22 m/s2 for 1 second and maintains uniform speed thereafter. 6. If we use the displacement formula s = ut + 1/2 at2; I arrived at 11.11 m. That is object moved the distance of 11.11m after 1 second. It is really confusing me. Because the Vehicle is moving at the uniform velocity of 22.22 m/s. So, the object should also moved at the same distance as Vehicle moved. But my calculation gives only 11.11m instead of 22.22 m/s. Where I am wrong? Can you please point out? thanks, pmk   Apr 1st 2018, 03:17 PM   #2
Senior Member

Join Date: Aug 2010
Posts: 434 Originally Posted by SpeedEC Hi Woody, I want to calculate the required force to move the object (70 Kg) for 1 sec at 22.22 m/s2. What is the distance traveled by the object for 1 sec at the same acceleration? 1. A Vehicle running at uniform velocity of 80 KMPH (50 MPH) i.e 22.22 m/s. 2. No acceleration since uniform velocity. 3. It picks the object while running. 4. What would be the acceleration of the Object? = 22.22 m/s2 ? as per the calculation that I mentioned above.
I don't understand exactly what is happening here. If the vehicle was running at 80 kmph and "picks" up the object without slowing down, then the object is instantaneously accelerated to 22.22 m/s. Of course that's impossible- the vehicle picking up the object must take some length of time to pick up the object.

 The object should accelerate at 22.22 m/s2 for 1 second and maintains uniform speed thereafter.
So you are assuming that the vehicle takes 1 second to pick up the object? What happens to the vehicle during that time?

 6. If we use the displacement formula s = ut + 1/2 at2; I arrived at 11.11 m. That is object moved the distance of 11.11m after 1 second.
If the object started from 0 speed and accelerated for 1 second at 22.22 m/s^2, it will have moved 22.22/2= 11.11 m in that second.

 It is really confusing me. Because the Vehicle is moving at the uniform velocity of 22.22 m/s. So, the object should also moved at the same distance as Vehicle moved. But my calculation gives only 11.11m instead of 22.22 m/s. Where I am wrong? Can you please point out? thanks, pmk
Your error is in thinking that the object was instantly going at the same speed as the vehicle when you have assumed that it takes one second to bring the object to that speed. Of course, it is physically impossible for the object to instantaneously jump to 22.22 m/s but you have not given enough information about exactly how the vehicle picks up the object to be able to determine how long it takes for the object to reach the vehicles speed.   Apr 1st 2018, 09:26 PM   #3
Junior Member

Join Date: Mar 2018
Posts: 12
Thanks HallsofIvy.

 I don't understand exactly what is happening here. If the vehicle was running at 80 kmph and "picks" up the object without slowing down, then the object is instantaneously accelerated to 22.22 m/s. Of course that's impossible- the vehicle picking up the object must take some length of time to pick up the object.
The object weight is 70 Kg.
The Vehicle weight is 45 tons (45000 Kg) and picks the object while it is running at uniform velocity of 80 KMPH (22.22 m/s).

Can you please tell me
1. the object's acceleration and
2. the time taken to reach the uniform speed the of the vehicle?
3. Force acted on an object while picking

Hope, reduction in speed of the Vehicle is negligible.

thanks
pmk   Apr 2nd 2018, 04:07 AM #4 Senior Member   Join Date: Jun 2016 Location: England Posts: 1,003 What this amounts to is how stretchy is the device used to pick up the package? During the acceleration of the package, it is moving slower than the vehicle, so it will fall behind the vehicle. In your initial post you mentioned 1 second to get to full speed, Thus the package accelerates at 22.22 m/s/s As you calculated, the package moves 11.11 m in this time. The Vehicle will have moved 22.22 m in the same time, so the package will be 11.11 m behind the vehicle! The force is easy to calculate from F=mA or 40*22.22 = 888.8 Newtons. I think what you are looking for is to balance the force required against the distance the pickup mechanism will have to stretch. Basically, shorter time, less stretch = more acceleration, more force. __________________ ~\o/~   Apr 2nd 2018, 10:56 AM   #5
Junior Member

Join Date: Mar 2018
Posts: 12
 The force is easy to calculate from F=mA or 40*22.22 = 888.8 Newtons.
40?
It is 70 Kg. F = ma; 70 x 22.22 = 1555.4 Newtons.

 I think what you are looking for is to balance the force required against the distance the pickup mechanism will have to stretch.
Exactly !!!

But the stretching distance that I can provide is only up to 2.22 meter. So, I want to know the impact on the object while picking the object when no shock absorber or stretching mechanism used. Then only I can able to design the stretching mechanism and other shock absorber for reducing the impact as possible.

can you help me?

thanks,
pmk   Apr 2nd 2018, 11:40 AM #6 Junior Member   Join Date: Mar 2018 Posts: 12 Case 1: No cushion or Shock absorber or stretching mechanism used around the Object 1. Vehicle running at uniform speed of 80 KMPH (22.22 m/s) 2. Pick the Object while running 3. So, Object should accelerate at 44.44 m/s2 in order to reach the same uniform velocity (22.22 m/s) of the Vehicle after 1 second. i.e. displacement should be same as the Vehicle's displacement of 22.22 m; so, required displacement (s) = 22.22 m; time (t) = 1 second; s = ut + 1/2 at2 = (0x0) + (0.5 x a x (1^2)); 22.22 = (0 + 0.5xa); 22.22 = 0.5a; thus Acceleration a = 22.22/0.5 = 44.44 m/s2 44.44 m/s^2 / 9.81 m/s^2 = 4.53 g So, In order to reach the uniform velocity of the vehicle, the object should accelerate at 44.44 m/s^2 (4.53 g-force) for 1 second NOT 22.22 m/s^2 as I said earlier. If the object accelerate at 22.22 m/s^2, the displacement will be only 11.11m and will be behind the vehicle as you said. Case: 2 We can reduce the g-force & impact on the object using cushion or Shock absorber or stretching mechanism around the Object. Am I right? I have to calculate required force for stretching mechanism and Cushion for distributing the load uniformly across the object. Any suggestions? thanks, pmk Last edited by SpeedEC; Apr 2nd 2018 at 11:49 AM.   Apr 2nd 2018, 05:43 PM #7 Senior Member   Join Date: Jun 2016 Location: England Posts: 1,003 Sorry, all sorts of mistakes in my post! However, you cannot pick up the object with no stretching, that would involve infinite acceleration. There are two equations you need to consider; s = ut + 1/2 at^2 v=u+at the initial speed "u" is zero so: s = 1/2 at^2 v=at you also need to consider the displacement of the vehicle: d=22.22t thus if you have s=22.22m after 1 second (a=44.44m/s/s) then v will be 44.44m/s after 1 second. so twice the speed of the vehicle! My calculations had the speed the same after 1 second, (so the displacement was 11.11m). Note that you can't just consider the end points. After 1/2 second at 44.44m/s/s, the speed of the package will be 22.22m/s but the displacement will be only 2.7775m the displacement of the vehicle in half a second will be 22.22*0.5=11.11m You might have to consider the rate of change of acceleration, which is called jerk (honestly, look it up:) __________________ ~\o/~ Last edited by Woody; Apr 2nd 2018 at 05:52 PM.   Apr 3rd 2018, 08:42 AM #8 Junior Member   Join Date: Mar 2018 Posts: 12 Dear all, I am so confused and also confused you !!! For your clarity, I am giving you the information as simple as possible to understand my requirement better. Object weight : 70 Kg Vehicle weight : 45000 Kg Object is a Solid . No stretching properties. Vehicle running uniformly at 80 KMPH Vehicle picks the Object while running uniform speed of 80 KMPH (22.22 m/s). After the Vehicle picking up the Object, the Object will maintain the speed of the Vehicle. My requirement: Is there any acceleration on an Object? If yes, What could be the acceleration of the Object? I want to design and apply proper stretching mechanism to reduce the impact on the Object. Hope you understand my requirement clearly. Kindly provide your valuable suggestions. thanks, pmk   Apr 3rd 2018, 10:18 AM   #9
Senior Member

Join Date: Oct 2017
Location: Glasgow
Posts: 444
 It is really confusing me. Because the Vehicle is moving at the uniform velocity of 22.22 m/s. So, the object should also moved at the same distance as Vehicle moved. But my calculation gives only 11.11m instead of 22.22 m/s. Where I am wrong? Can you please point out? thanks, pmk
An object travelling at 22.22 m/s for one second is going to cover more distance than an object travelling at variable speed between 0 m/s and 22.22 m/s (at constant acceleration). This is normal.

Perhaps what you want to model is something different... something a bit more akin to somebody "picking up" an object and putting it into their car... or a crash. Let's try going through your points and figure things out in a different way.

 Originally Posted by SpeedEC 1. A Vehicle running at uniform velocity of 80 KMPH (50 MPH) i.e 22.22 m/s.
So we have

$\displaystyle u_{car} = 22.22$ m/s
$\displaystyle u_{obj} = 0$ m/s
$\displaystyle m_{car} = ?$ kg
$\displaystyle m_{obj} = 70$ kg

So we have some information relevant for a collision problem.

 2. No acceleration since uniform velocity.
This is okay. In collision problems, we can assume that the collision is instantaneous and we just consider conservation of momentum before and after the collision.

 3. It picks the object while running.
So we can use the equation of the conservation of momentum:

$\displaystyle m_{car} u_{car} + m_{obj} u_{obj} = (m_{car} + m_{obj}) v$

Consequently, we have

$\displaystyle 22.22 m_{car} = (m_{car} + 70) v$

Solving for v gives:

$\displaystyle v = \frac{22.22 m_{car}}{70 + m_{car}}$

No meaningful solution exists for $\displaystyle v >= 22.22$. Therefore, if the mass of the car is something like 45000 kg, we have

$\displaystyle v = \frac{22.22 \times 45000}{70 + 45000} = 22.185$ m/s

So, in this scenario, the interaction is modelled as if the car picks up the mass and then has to slow down because of the conservation of momentum. I guess this isn't really the kind of information you want though.

If we consider a momentum exchange over a time period (the collision time), then we can use the momentum exchange to find out the force exerted by the object on the car. Let's say we consider a 45000 kg car. The momentum exchange is:

$\displaystyle \Delta p = m_{car} (u_{car} - v)$
$\displaystyle = 45000 \times (22.22 - 22.185) = 1553$ kg m/s

And using Newton's second law (note: F=ma is the law that can be derived from the rate of change of momentum, which is the actual definition of Newton's second law) with a collision time is one second, we get

$\displaystyle F = \frac{\Delta p}{\Delta t} = \frac{1553}{1} = 1553$ N

That force is rather low, but considering that the collision time is rather long, it makes sense. Note that a collision time of 1 second is much longer than typical collision times of 0.05 to 0.001 seconds. Perhaps the car has a crumple zone Finally, the acceleration is:

$\displaystyle a = \frac{F}{m_{obj}} = \frac{1553}{70} = 22.186$ m/s$\displaystyle ^2$

If you absolutely require the acceleration to be 22.22 m/s$\displaystyle ^2$, you will either have to change the collision time or the initial velocity of the car.

Last edited by benit13; Apr 3rd 2018 at 10:28 AM.   Apr 4th 2018, 01:28 AM   #10
Junior Member

Join Date: Mar 2018
Posts: 12
Hi benit13,

Thanks for giving valuable data. Finally, I think almost, I have reached the correct place where I actually want to be.

Force on the Object (F) = 1553 Kg-m/s
Acceleration of the Object (a) = 22.186 m/s2 = 2.26 g-force

 If you absolutely require the acceleration to be 22.22 m/s2, you will either have to change the collision time or the initial velocity of the car.
This is the place where I have to consider to design the stretching mechanism in order to reduce the impact on the Object.

 you will either have to change the initial velocity of the car.
Not possible.

 you will either have to change the collision time
Yes. I have to use stretching mechanism in order to reduce the impact on the object from the Vehicle's Force. For example, providing extension springs in the Vehicle in order to give cushion to the object to protect it from the 1553 Kg-m/s sudden force while picking the Object. I need to calculate the time to provide the cushion using spring for that. Any idea?

Vehicle will continue its speed at 80KMPH approx (22.185 m/s) for few seconds. My ultimate requirement is to protect the Object from the force of the Vehicle using Stretching and Other Mechanisms.

great thanks,
pmk

Last edited by SpeedEC; Apr 4th 2018 at 01:36 AM.  Tags force, move, object, required Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post JessicaO Thermodynamics and Fluid Mechanics 4 Dec 16th 2014 12:59 PM jlyu002 Thermodynamics and Fluid Mechanics 2 Aug 13th 2014 06:44 AM jlyu002 Kinematics and Dynamics 3 Jul 12th 2014 10:36 AM bl71236 Kinematics and Dynamics 5 Aug 9th 2011 11:53 AM sentientnz Energy and Work 1 Aug 1st 2009 02:30 AM