Calculate Min Force to Get Block Moving: FORCE PROBLEM

In summary, the conversation discusses a problem involving a participant pushing a block of ice on a frozen lake with a certain force and the calculation of minimum force, acceleration, and time needed to reach a top speed. The conversation covers the free body diagram, components of the force, and the equations needed for solving the problem. The correct formula for minimum force is F= (0.1(50x9.8))/(cos25 - (0.1sin25)), which gives a value of 56.01N. The formula for acceleration is a=(Fappcos/theta-ukFn)/m, and the value of Fapp is the applied force.
  • #1
gianne671
16
0
μ

Homework Statement


Hi I am 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:
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  • #2


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


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


Yes. But you have to think about the other forces which also act on the block.
 
  • #5


ohh isee! I think there are two forces. it's the normal force and the g.. :) is that it?
 
  • #6


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


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


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


Im trying to draw it.. but i can't still get it.. :( I am sorry for being a NOOB.. (teary eyes)
 
  • #10


here is the diagram
 

Attachments

  • diagram.doc
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  • #11


tnx!

so if I am 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


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


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.. I am really sad right now.
 
  • #14


gianne671 said:
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


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


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


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


gianne671 said:
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


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


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


I bet Fapp is the applied Force.
 
  • #22


I am so sorry!

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

FAQ: Calculate Min Force to Get Block Moving: FORCE PROBLEM

What is the force required to get a block moving?

The force required to get a block moving is known as the minimum force. It is the smallest amount of force needed to overcome the static friction between the block and the surface it is resting on.

What is the formula for calculating minimum force?

The formula for calculating minimum force is F = μs * N, where F is the minimum force, μs is the coefficient of static friction, and N is the normal force acting on the block.

How do I determine the coefficient of static friction?

The coefficient of static friction can be determined by conducting an experiment where the force required to move the block is measured and compared to the normal force acting on the block. The coefficient can then be calculated using the formula μs = F/N.

What factors can affect the minimum force required?

The minimum force required to move a block can be affected by a variety of factors such as the weight of the block, the surface it is resting on, and the roughness of the surfaces in contact. The coefficient of static friction can also vary depending on the materials involved.

How can calculating minimum force be useful in real-world applications?

Calculating minimum force is useful in real-world applications such as designing machines or structures that involve sliding or rolling objects. It can also be used in determining the maximum weight a person or machine can push or pull without exceeding the minimum force required.

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