Determining Maximum Torque in a Truck and Forklift System

[scatterbrainz]

Homework Statement

not a homework question - just a physics questions

Relevant Equations

F = ma, v=v+at

This maybe complex and complicated - I know its going to be multi step. Please see attached picture. It involves velocity and weight and gravity torque force, pivot points.

 

[scatterbrainz]

I'm not even sure which equations to start with F = m x a, but convert a vector to an acceleration, but is it acceleration or velocity since at the point where the front forklift axle meets the trailer floor, it has to be convert to Force (N). Is there not F(gravity) also playing a role in there. Friction and air resistance = 0. Thanks for help

 

[berkeman]

Welcome to PF.

The figure is too low resolution to get much information from it. Can you say in words what it is asking? Or maybe link to a higher resolution version of the image?

 

[kuruman]

Whatever the question may be, it is obvious from the picture that the end of the incline should have been placed on the bed of the truck at a point horizontally between the rear and front wheels.

 

[haruspex]

I'm not even sure which equations to start with F = m x a, but convert a vector to an acceleration, but is it acceleration or velocity since at the point where the front forklift axle meets the trailer floor, it has to be convert to Force (N). Is there not F(gravity) also playing a role in there. Friction and air resistance = 0. Thanks for help

Which acceleration do you think is relevant and why?

 

[scatterbrainz]

You have forklift of "x" weight. which is greater than the weight of the truck. Since the forklift has momentum - it generates a "force". So in my head - it would have acceleration on both an x-axis and y-axis since it is vector. The time length to travel the distance might be 3 seconds. if I added the x-axis and y-axis velocities - I think I would get the velocity of the vector. That velocity turns into force at impact point "X" - which is about 6 inches past where the dock plate ends and 78 inches behind the pivot point. Creating a teeter-totter affect. Once side absorbs the force of the front axle of the fork lift. where the front of the truck is torqued upward off the ground due to the impact. Part of the question becomes, how much of that gross forklift weight of 16,357.4 lb is at the centre of gravity of the forklift, because when you put a load onto the forks, the centre of gravity moves forward. Please take note of 1 error: distance between truck axles is not 270 inches rather 156 inches, since the total box is 240 inches. 84 inch overhang behind rear axle. Thanks for assistance.

 

[haruspex]

Since the forklift has momentum - it generates a "force".

I see that the forklift comes off the ramp at 0.4m/s, rather than slowing to a stop there. Yes, that means it has momentum, but the change from ramp to truckbed is sudden, so you cannot say what the peak force is. You can say it exerts an impulse, and you can calculate that as the 'instantaneous' change in momentum of the forklift. To calculate that, you need to determine its velocity vector just after touching the truckbed. Which way is its mass centre moving then?

A consequence of the impulse is that the truck will tilt a bit. Whether that results in the front tyres leaving the ground depends on the suspension and tyre pressure.

That vector will continue to change as the forklift moves further onto the truck. This implies a centripetal force supplied by the normal force from the truckbed.

 

[scatterbrainz]

The entire mass of the forklift & load is always moving in a forward direction. What I am looking for is that moment where the wheel leaves the plate and its full weight as realized by the forklift's centre of gravity is on the truck floor. for that instantaneous moment the weight on the axle will be much greater than the rest of the truck weight. The suspension is air ride - and in a deflated state - the tires are standardized pressure for the tire. It is direct transference to the ground via the truck floor to the frame (chassis) to the 2 axles (tires). SO what I'm trying to determine is the following the maximum weight the forklift and payload can be before the front lifts off of the ground. and When this tolerance is breached, given this example, if the front axle will come off the ground 6 inches, 12 inches, or more? If I put 5 Newtons of force on 1 side of a teeter-totter - the otherside will go up by 5 Newtons less whatever may be weighing it down to begin with, so 5N Left transfer to 5N right minus (lets say 2 lbs weighing it down already). that means the right side goes up by 3 Newton. I just need a way of seeing it via formulas.

 

[haruspex]

that moment where the wheel leaves the plate and its full weight as realized by the forklift's centre of gravity is on the truck floor. for that instantaneous moment the weight on the axle will be much greater than the rest of the truck weight

The weight transference is more gradual than that. As the forklift moves down the ramp, the normal force between truck and ramp steadily increases. Even when the front wheels reach the truck, the rear wheels are on the ramp, so the truck is not taking quite the full weight. The transference continues smoothly until the forklift is fully on the truck.

However, as I posted, there is a sudden impulse when the front wheels leave the ramp because there is a sudden change of direction of the velocity of the forklift’s mass centre.
Try to find an expression for that impulse.

that means the right side goes up by 3 Newton.

I do not know what you mean by that. 3 Newtons is not a distance.

 

[berkeman]

I do not know what you mean by that. 3 Newtons is not a distance.

I'm guessing they are referring to Newton's 3rd Law, but who knows. Look at their username...

 

[scatterbrainz]

IF i have a weight of kg on the right side of a teeter totter and I apply weight of 10 kg on the left side, then the left side = F = m x a = 10 x 9.81 m/s x 1s = 98.1 N (kg x m/s2). Since the right side has 3 kg already, then the right side is F = m x a = 3 kg x 9.81 m/s x 1 s = 29.41 N. SO I have 98.1N on the left and 29.41 N on the right - 98.1 - 29.41 = 68.89 N - which means the right side will travel will be displaced by 68.89 N which is also a vector on the y.

 

[haruspex]

IF i have a weight of kg

I guess you meant 3kg

F = m x a

Well, F = mg.

m x a = 10 x 9.81 m/s x 1s

Where does the 1s come from? And 10 x 9.81 m/s x 1s=98.1m. You mean 10 x 9.81 m/s².

98.1 - 29.41 = 68.89 N

I get 68.69, but applying the usual precision rules, 68.7.
But why are you subtracting one of these forces from the other? They act in the same direction, so add to produce a force 127.5N on the fulcrum.
To determine whether the teeter totter rotates you need to calculate the torques, force times distance.

the right side will travel will be displaced by 68.89 N

68.89N is neither a displacement nor a distance travelled. If you calculate the torque difference then you could determine the angular acceleration.

In the context of the truck and forklift, you need to determine when the forklift exerts the maximum torque about the truck's rear wheels. That is not trivial. While the forklift advances down the ramp it is straightforward because the normal force between the ramp and truck increases; but after the front wheels of the forklift come off the ramp it is more complicated. Although the total force (ramp+forklift front wheels) continues to increase a little, the distance of that force from the truck rear axle decreases. When does the torque reach a maximum?