Average force exerted on the ball

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SUMMARY

The discussion centers on calculating the average force exerted on a ball during a kick, specifically addressing two parts: the average total force and the force exerted by the foot. The equations used include F = (mv2 - mv1) / Δt for both parts, with the second part requiring the subtraction of gravitational force (mg). The final calculated forces are 45 N for the average total force and approximately 48.87 N for the force exerted by the foot, factoring in the direction of the forces involved.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Familiarity with vector addition and subtraction of forces
  • Knowledge of kinematic equations related to motion
  • Basic grasp of gravitational force and its impact on motion
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  • Study vector decomposition in physics to better understand force directionality
  • Learn about kinematic equations in detail, focusing on their applications in real-world scenarios
  • Explore the concept of average force in different contexts, such as sports physics
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nysnacc
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Homework Statement


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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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