Average force exerted on the ball

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Homework Help Overview

The discussion revolves around calculating the average force exerted on a ball, with a focus on separating forces in different directions and understanding the implications of gravity on these forces. The subject area includes dynamics and vector analysis in physics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the separation of velocity components in the x and y directions and question the relationship between different parts of the problem. There are discussions about the equations to use for calculating average force and the impact of gravity on these calculations.

Discussion Status

The conversation includes attempts to clarify the equations for average force and the distinction between total force and the force exerted by the foot. Some participants express uncertainty about the signs of forces and the implications of the angle provided in the problem.

Contextual Notes

Participants note the importance of considering vector directions when calculating forces and question the assumptions made regarding the weight of the ball and its effect on the total force required to achieve a certain velocity.

nysnacc
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Homework Statement


upload_2016-10-12_10-43-9.png


upload_2016-10-12_10-43-34.png

upload_2016-10-12_10-43-55.png


Homework Equations


∑Fav = mv2 - mv1

The Attempt at a Solution


We have to separate v in x dir and y dir?
 
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nysnacc said:
We have to separate v in x dir and y dir?
Why? (Note that it starts from rest.)
 
but are part one and part two asking the same thing?
 
nysnacc said:
but are part one and part two asking the same thing?
No. One asks for the average total force on the ball. The other part asks for just the force exerted by the foot.
 
For the average total force on the ball (1st part). what equation should i use?

For the (2nd part) force exerted by the foot, the equation is F = (mv2- mv1 ) / Δt
 
nysnacc said:
For the (2nd part) force exerted by the foot, the equation is F = (mv2- mv1 ) / Δt
No, that will give the average total force on the ball.
 
That's total force exerted on the ball?
 
For part 1,

F = 1/ Δt * ∫ F dt ?? which is (mv2- mv1 ) / Δt ??

For part 2,

F = (mv2- mv1 ) / Δt - mg (cuz the question said subtract the weight)
 
nysnacc said:
That's total force exerted on the ball?
Yep.
 
  • #10
nysnacc said:
For part 1,

F = 1/ Δt * ∫ F dt ?? which is (mv2- mv1 ) / Δt ??
Yes.

nysnacc said:
For part 2,

F = (mv2- mv1 ) / Δt - mg (cuz the question said subtract the weight)
Careful. Forces are vectors, so here you need to consider the directions of each.
 
  • #11
For part 2

(mv2- mv1 ) / Δt (going up)

mg (going down)

what should i do with the 60 deg which is given?
 
  • #12
nysnacc said:
what should i do with the 60 deg which is given?
That's the direction of the average total force on the ball.
 
  • #13
Part 1, F = 45 N
Part 2 F = 41.236 N?
 
  • #14
nysnacc said:
Part 1, F = 45 N
Part 2 F = 41.236 N?
I'm not sure whether the hint given in the question is simply wrong or is assuming the weight, being a downward force, will be expressed as a negative quantity. Ask yourself this: does the presence of gravity increase or reduce the force needed from the boot to achieve the given velocity?
Secondly, forces are vectors. You cannot add (or subtract) forces acting in different directions by adding their magnitudes.
 
  • #15
Is #1 correct, 45 N
 
  • #16
#2 i have F = Fav (cos i + sin j) + mg (-j)

= (Fav cos )i + ( Fav sin - mg ) j
 
  • #17
nysnacc said:
Is #1 correct, 45 N
Yes.
 
  • #18
Doc Al said:
Yes.
why is my #2 not correct?
 
  • #19
nysnacc said:
why is my #2 not correct?
You made a mistake with the sign of the weight vector.
 
  • #20
Fav - mg (-j) ??
so it is a plus?
 
  • #21
So 48.87 N?

#2 is greater than #1
 
  • #22
##\vec{F_{total}} = \vec{F_{kick}} + \vec{F_{weight}}##
To find the force of the kick you must subtract the weight vector, but the weight points down.
 
  • #23
nysnacc said:
So 48.87 N?

#2 is greater than #1
Yes.
 

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