FORCE PROBLEM. how calculate the min. force to get the block moving, given μk and μs.

  • Thread starter gianne671
  • Start date
  • #1
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μ

Homework Statement


Hi Im a new user here.. i have problem with my physics homework... can you help me? Like this one.. I'm really really stuck for about 3 hours..

1. A participant in a winter games event pushes a 50-kg block of ice across a frozen ice lake with a force F directed at 25 degree below the horizontal. if the coefficient of kinetic and static friction are 0.03 and 0.1 respectively,

a) calculate the minimum force to get the block moving
b) calculate the acceleration of the block if that force is maintained.
c) how long dies it take the contestants to reach a top speed of 5.0m/s?


Homework Equations


F=ma
a=Fnet/m



The Attempt at a Solution


I have drawed the free body diagram of this problem..
 
Last edited:

Answers and Replies

  • #2
998
15


What forces act on the block for part (a)?
 
  • #3
16
0


the force acting on the block was named force "F" directed at 25 degree below the horizontal..
 
  • #4
998
15


Yes. But you have to think about the other forces which also act on the block.
 
  • #5
16
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ohh isee! I think there are two forces. it's the normal force and the g.. :) is that it?
 
  • #6
998
15


Yes. the normal force of surface on block and the pull of the Earth on block.

Now what are the components of F in the horizontal and in the vertical?
 
  • #7
16
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Fx= kinetic friction of 0.03 and static of 0.01
Fy= normal force and the pull of earth on the block and 25 degree?

that's all I can think of ..
 
  • #8
998
15


That is not what I meant.

Consider F, the force with which the participant is pushing the block at 25 deg below the horizontal. What are the components of THIS F in the vertical and in the horizontal?
 
  • #9
16
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Im trying to draw it.. but i can't still get it.. :( im sorry for being a NOOB.. (teary eyes)
 
  • #10
998
15


here is the diagram
 

Attachments

  • diagram.doc
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  • #11
16
0


tnx!

so if im not mistaken.. Fappcos/theta-f=maFn+Fappsin/theta=Fappcos/theta-u..
then..
Fappcos/theta-mukFn=maFn=Fg-Fappsin/theta
that gives me..

a=Fappcos/theta-ukFn/m
am i right?
 
  • #12
998
15


But for part (a) the block is just about to start moving. Hence we can assume it is at rest. Hence there is no acceleration.
 
  • #13
16
0


I see.. so is this the right formula?

Fcos/theta-us(fg-Fsin/theta=0
Fcos/theta-usFg+usFsin/theta=0
Fcos/theta+usFsin/theta=usFg
F=us x Fg/cos/theta+ussin/theta

so..
F= (0.1(50x9.8))/cos25+(0.03(sin25))

is that it????

if not.. can you please illustrate the formula to be used for A) B) & C) because honestly, I can't do it anymore.. Im really sad right now.
 
  • #14
998
15


I see.. so is this the right formula?

Fcos/theta-us(fg-Fsin/theta=0
Fcos/theta-usFg+usFsin/theta=0
Fcos/theta+usFsin/theta=usFg
F=us x Fg/cos/theta+ussin/theta

so..
F= (0.1(50x9.8))/cos25+(0.03(sin25))

is that it????

Almost correct! But in the first equation the minus in brackets must be plus because the normal force must be equal and opposite to (Fsintheta + mg).

Hence Fcos/theta-us(fg + Fsin/theta)=0
 
  • #15
16
0


so the only problem is my first equation?
then the rest 2nd,3rd,4th and Final equations are correct? :)

can you please show me how to compute the time? needed to achieve a 5m/s?
 
  • #16
998
15


Fcos/theta-us(fg + Fsin/theta)=0
Fcos/theta-usFg - usFsin/theta=0
Fcos/theta - usFsin/theta=usFg
F=us x Fg/(cos/theta - ussin/theta)

so..
F= (0.1(50x9.8))/(cos25 - (0.1sin25))
 
  • #17
16
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what did you get?
56.70N?

then for the acceleration the formula is = a=Fappcos/theta-ukFn/m am I right? what nos. did it give you(assuming that u already substituted for their value,, so that I can compare)
and how can I solve for the time?
 
  • #18
998
15


what did you get?
56.70N?

then for the acceleration the formula is = a=Fappcos/theta-ukFn/m am I right?

I got 56.01N.

You better put a bracket as shown below:

a=(Fappcos/theta-ukFn)/m

Your Physics is OK.
 
  • #19
16
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hey can i ask? what is the Fapp in the formula of acceleration???

and why do I keep getting 56.70N instead of 56.01???

i got this value of 49/0.8641=56.70N
 
  • #20
16
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so the value which i get in the force needed to keep moving is 56.01 is the force app?? so..

56.01cos 25-0.03*?/50 HELP!
 
  • #21
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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1,043


I bet Fapp is the applied Force.
 
  • #22
998
15


I am so sorry!

I meant that I got 56.71N. Your calculation is OK!
 

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