Calculating Min. Stopping Dist. of Lorry Carrying Crate w/ Coef. of 0.3

[johnny_702]

hey people i read a post similar to my problem but just couldn't get it... this is my problemThe coefficient of static friction between the
flat bed of the lorry and the crate it carries is
0.30 Determine the minimum stopping
distance s that the lorry can have from a
speed of 70 km/h with constant deceleration
if the crate is not to slip forward.

there is a free body diagram showing that the crate is 3m away from the front of the flatbed lorry and the answer is said to be 64.3m but i can't seem to get it.
thanks!

 

[tiny-tim]

Welcome to PF!

Hi johnny_702! Welcome to PF!

(if "the crate is not to slip forward", then its distance from the front is irrelevant )

What is the maximum force on the crate if it doesn't slip?

So what is the maximum possible deceleration?

Then use one of the standard constant acceleration equations to find the stopping distance of the lorry …

what do you get? ​

 

[Meithan]

As the lorry brakes, it undergoes a certain acceleration $$\vec{a}$$ directed backwards. This means the surface of the flatbed wants to slide backwards under the crate. Friction tries to prevent this, by making the crate decelerate at the same rate as the lorry. However, this only happens up to a certain point where friction becomes insufficient. The maximum force or acceleration that friction provides depends on the static friction coefficient, $$\mu_s$$.

So in order to solve this problem, you have to answer the following points:

1) What is the acceleration the lorry undergoes as it brakes from 70 km/h to a full stop? Your answer will depend on the initial speed of the lorry $$v_i$$ and the braking distance $$d$$ (which we don't know yet).

2) What is the force on the crate provided by friction? Your answer will depend on the mass of the crate $$m$$ (which we don't know yet), the static friction coefficient $$\mu_s$$ and the acceleration of gravity $$g$$. Note that this force is the maximum force that friction can provide.

3) Applying Newton's Second Law to the previous result, to obtain the acceleration of the crate produced by friction. You'll notice the mass of the crate cancels out (and so, we don't need to know it). This acceleration is the maximum acceleration friction can provide.

In 1) we calculated the actual acceleration the lorry undergoes. In 3) we obtained the maximum acceleration friction can provide, thus preventing the crate from sliding. If these two are equal, we are on the limit. Make the lorry's acceleration just a little larger, and friction won't be enough to keep the crate from sliding. So:

4) Set these accelerations equal to each other, and find the limiting value of $$d$$, the braking distance. This is a minimum.

Hope this helps.

 

[johnny_702]

hey thankyou tiny-tim! I can't believe i have never found this place before. I just realized I had messed up the value for the coefficient. edited it there. thanks for the help guys I am going to work through it now.

Just one question though to Meithan, how can I work out the force provided with friction without the mass of the crate? I know you state we don't need it in the next point but could you explain how it cancels out please?

Thanks again guys its right about now I wish I had paid more attention in my lectures :)

 

[tiny-tim]

Thanks again guys its right about now I wish I had paid more attention in my lectures :)

he he!

Just one question though to Meithan, how can I work out the force provided with friction without the mass of the crate? I know you state we don't need it in the next point but could you explain how it cancels out please?

For the force you do need the mass, but for the acceleration, you'll find the mass cancels out.

 

[johnny_702]

Yea but how do I work out the mass? F =ma but I only have acceleration so I can't work out the force you see? at least I think. Does it all work out for you guys when you do it? Maybe I've been looking at this too long and my brains wrecked :/

 

[tiny-tim]

Call the mass m …

then what's the friction force? ​

 

[Meithan]

As tiny-tim says, just call the mass 'm'. When I ask the force, you don't need to give a number -maybe that's what's bugging you-, just the mathematical expression, the formula.

For example, if I told you an object experiences a force 'F' and has mass 'm', you could write that its acceleration is a = F/m - and it doesn't matter what the actual numbers are, the formula is always valid.

I know it's kind of weird to think this way. It's all just symbols! But with practice you'll get used to this abstraction. And in physics it's always preferable to give an answer with symbols than with a numerical answer.

So, going back to your problem. What is the formula for the force of friction on the crate?

 

[johnny_702]

ahh right right i see. And is this the same when working out your point 1?
em force of friction is the coefficient of static friction*force acting on the crate?

which is 0.3*(m*9.81)= 2.943m?

 

[johnny_702]

by the way people when I said the answer was 64.3m it doesn't tell you this in the question. My lecturer just put it at teh end so you would know if you did it right or not
thanks

 

[tiny-tim]

em force of friction is the coefficient of static friction*force acting on the crate?

which is 0.3*(m*9.81)= 2.943m?

Yes, that's the maximum possible static friction force.

So the maximum possible deceleration is … ?

 

[johnny_702]

f=ma so 2.943m/m= 2.943m/s?

I think I might be getting the hang of this...(unless i messed it up there lol)

 

[johnny_702]

YES! I got it there I think.

If the max acceleration for the crate is 2.943 then I plugged it into the equation

v^2=u^2+2as

0=380.25+2(2.943s) therefore 380/2=2as, as=190, 190/a=s.190/2.943=64.6m

Just a we bit off but that's probably because of my accuracy rounding up etc...
thanks very much guys! :) Thought id never get this one

thanks again your help is very much appreciated. Just took me a while to catch on to what you had been telling me from the start lol