# Homework Help: Motion near earths surface

1. Jul 30, 2007

Ok i tried these questions over and over and i cannot seem to grasp the idea so if someone could please just give me the full answer to this it would be greatly appreciated so i can see fully how to do it and i can apply it to similar questions...

1. a) Cathy takes the bus home from work. In her hand she holds a 2.0 kg cake box, tied together with string. As she ascents the steps into the bus, the box accelrates upward at a rate of 2.5 m/s^2. What is the force exerted on the string? (ok is this a tention question or how do i do this, can anyone prove an explanation or a full answer? )

b) Cathy sets the box on the seat beside her. the bus accelrates from rest to 60 kn/h in 4.0s, and the box begins to slide. What is the coeffiecent of static friction between the box and the seat of the bus? (ok i know that Ff=Us(Fn) I know Fn = ma i know a=v2^2-v1^2/2d i found d by doing .5(v1 + v2)(t) so i can find accelration i find Fn = m which is 2 kg and acelartion which i believe is 4.1675 m/s^2 but i am left with a equation iwht 2 variables Ff=Us(Fn), Therfore Us = Ff/Fn i know Fn but i do not know Ff, please someone enlighten me)

c)A taxi suddnelt cuts in front of the bus, causing the bus driver to slam on the brakes. The bus driver reduces speed from 60km/h to 20km/h in 1.5s. Does Cathy's cake slide forward? (is this Uk??? i need major help with this one, this is accelration isnt it?)

Ok Next question...I have to design and complete an investigation to determine how the folloing factors affect the force of friction:
wieght
Surface area
speed
I need to measure the magnitude of the force of friction. any suggestions on how i should go about this?

2. Jul 30, 2007

### mgb_phys

Do you know f = m a ?

F = m a again, you are then supposed to assume that this is just enough to overcome friction. You know that friction foce = weight * coefficient.

f = m a again!! This time you also know the fricional force trying to stop it accelearating forward

You can measure the frictional force by simply measuring the force needed to start the object slipping ( using a spring balance?).
Then from knowing force = weight * friction coef.
Do study the other values rememebr that you want to vary only one parameter at once.

Last edited: Jul 31, 2007
3. Jul 30, 2007

i do know f = ma but the question cant be that simple A) plus it is being exerted upward so wouldnt it have to be Fn= ma - (-fg)?? thus Fg = 2(9.8) = 19.6N Fn= 2.0 (2.5)-(-19.6) = 24.6 N????this right ??

4. Jul 31, 2007

### mgb_phys

It is almost that simple, you have f = m a downward, where a=g and at the same time f = m a upward where a=2.5. Assuming you typed the figures into your calculator correctly! They obviously add because jerking it upward increases the force on the string, it's always worth trying to understand which direction the forces act rather than just hoping to get all the signs correct.

b, For the bus accelerating, first convert everything into m/s
Then you know the start speed =0, you know the end speed and you know time so 'a' shouldn't be too difficult - check out the equatons of motion in the physics formulae thread https://www.physicsforums.com/showthread.php?t=110015

You know the normal force down on the seat ( the weight), you can calculate the forward force ( f=ma) , and so the friction coef is easy.

5. Aug 2, 2007

ya but i need to get the static friction Ff=UsFn thats cool that i have Fn now but how doi get another variable ????

6. Aug 2, 2007

i need a lot of help with this this is getting me soo frusterated becuase i do not have another number i only know the Fn thats it FF=UsxFn i need another thing then just Fn how do i find Ff or Us???

7. Aug 2, 2007

### nrqed

I think that it is pedagogically more sound to always tell students that there is only one acceleration vector at any time for any object. It is my experience that talking about more than one acceleration is simply misleading and confusing. The equation simply says that the net force is equal to m times the (unique) acceleration.

8. Aug 2, 2007

9. Aug 2, 2007

### nrqed

I will start with this one since it's not totally clear to me what the situation is.

To do those problems, it is always very very useful to start with a free body diagram. Believe me, once you have the FBD, more than 3/4 of the work is done.

But I am not sure I can picture the situation here. Do they mean that she is holding a string which is attached to the box? In that case, what are the forces and their directions acting on the box?

10. Aug 2, 2007

serously i dont know i honestly do not know i take correspondants physics and they give me these books i have no teacher and no help except this site. thats the question i dont think she is holding the string but its very possible shes holding the string but logically who holds a cake with a string attachted to it like its your dog?? so im going to say that the string is just holding the package togeather? wat do u think?? thats a very valid point tho what is the question asking?

11. Aug 2, 2007

i seroisuly believe that the string is just holding the box togeather becuase it says she is holding the cake which is help togeather by string so i know that a force in the upward motion is Fn=ma-(-fg) becuase it is moving up ward thus fg = 2 (9.8) =19.6N then ma = 2(2.5) thus Fn =2(2.5)-(-19.6) = 24.6 N

Now B part is a little different i have v1=0.0m/s v2=60km x 1000 / 3600 = 16.67m/s i have t=4s and i need to find Ff=Us(Fn) ?? or is there another eq

12. Aug 2, 2007

as for a free body diagram...this is wat i believe it looks like please correct this if i am wrong before in question a it was going vertical but now if its on a seat then it will be horisontal Fapplied<----o->Ff ??? hte Fapplied is the force applied by the accelration correct?

13. Aug 2, 2007

### mgb_phys

Ok - looks like you have tried !!!!
a, There is a force downward of = 2kg * 9.8m/s^2
And a 'jerk' upward of 2kg * 4m/s^2
So total F = 2*9.8 + 2*4 = 24.6N
As I said, rather than just trusting getting the signs correct think about the forces, the weight pulls down and the acceleration upward both are trying to break the string so the overall force must be larger, ie add them

b, The bus accelarates from 0 - 60Km/h in 4s that is 4m/s^2 (assuming I did the sums correct!). By F=ma that exerts a sideways force on the 2Kg box of 8N.
The co-efficent of friction is defined so that the sideways force = the downward force * friction.
We know the downward force is F=2g ie 2*9.8 = 19.6N
The sideways force we just calculated as 8N so the friction.
F (pull) = F (normal) * friction, friction = Fpull/Fnormal = 8/19.6 = 0.41

c, The bus (de)accelarates from 60Km to 20Kmh, ie 40Km/h in 1.5s =7.33m/s^2
The force needed to overcome sideways friction we just calculated as 8N
The force due to (de)accelaration is F = m a = 2kg * 7.33 = 14.67N
This is larger than the friction (8N) so it slides!

ps - please check my actual numbers I was doing the sums in my head ;-)

14. Aug 2, 2007

how did u get the accelration to be 4? did u sue the forumula v2=v1+2a(t) thus v2-v1/2t 16.67m/s-0.0m/s / 2(4) thus equalling 2.01 m/s^2???? or how did u do it?

15. Aug 3, 2007

a=2.08m/s^2 Ff=2.0(2.08) =4.16N Fn=2(9.8) = 19.6N

{Ff=Fn(Us)} 4.16N=19.6(Us) Us=4.16/19.6 = 0.21 ???
CORRECT?

16. Aug 3, 2007