What Is The Maximum Force The Locust Exerted on The Plate?

  • #1
BurpHa
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Homework Statement:
The jumping ability of the African desert locust was measured by placing the insect on a force plate, a platform that can
accurately measure the force that acts on it. When the locust
jumped straight up, its acceleration was measured to follow the
curve in Figure P4.71. What was the maximum force that this
0.50 g locust exerted on the force plate?
Relevant Equations:
Newton's Second Law: F = ma.
At the beginning, I just looked for the highest point in the graph, which is approximately 90 \frac m s^2.
Then I plugged it in the formula F = ma and got the force equaled to 0.045 N.

However, when I looked back, the graph is about the change in acceleration. So really, I'm dealing with acceleration of acceleration. From what I understand, the acceleration increases when the locust jumps up until t is approximately about 27 ms, when it starts to decrease.

The answer I get from my book is 0.084 N. When I plug 0.084 N to the formula F = ma, the acceleration is 168 \frac m s^2

I don't understand how it got 168 \frac m s^2 so that it could get 0.084 N.

Thank you for your help.
 

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Answers and Replies

  • #2
PeroK
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The rate of change of acceleration is called jerk. But, that's not relevant here.

I'm not sure how the book got its answer, but you did forget something!
 
  • #3
BurpHa
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The rate of change of acceleration is called jerk. But, that's not relevant here.

I'm not sure how the book got its answer, but you did forget something!
Do you mean the gravitational force?
 
  • #5
BurpHa
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Do you mean the gravitational force?
Yes, I've been trying to understand how the book (College Physics: A Strategic Approach 4th edition by knight, jones, and field) got its answer
 
  • #6
PeroK
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Yes, I've been trying to understand how the book (College Physics: A Strategic Approach 4th edition by knight, jones, and field) got its answer
I can't help you there. I just googled for "speed of jumping locust" to corroborate their data. You couldn't do that when I was a student!
 
  • #7
BurpHa
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I can't help you there. I just googled for "speed of jumping locust" to corroborate their data. You couldn't do that when I was a student!
I also googled, but the answer I got was the one I got, 0.045 N.
 
  • #8
BurpHa
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I can't help you there. I just googled for "speed of jumping locust" to corroborate their data. You couldn't do that when I was a student!
But the thing is that the acceleration changes when the locust in in its motion. According to the graph, it looks like the acceleration increases when the locust jumps up.
 
  • #9
PeroK
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I also googled, but the answer I got was the one I got, 0.045 N.
For such a high acceleration, the extra force to overcome gravity is not as significant as it would normally be.
 
  • #10
BurpHa
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For such a high acceleration, the extra force to overcome gravity is not as significant as it would normally be.
So you agree the answer is 0.045 N? I mean, we don't have a fixed acceleration as we do in other problems. This time, the acceleration changes as the locust progresses through its jump.
 
  • #11
haruspex
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So you agree the answer is 0.045 N?
Plus the force to overcome gravity, which adds about 10%.
 
  • #12
BurpHa
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Plus the force to overcome gravity, which adds about 10%.
Do you think 0.084 N is the correct answer?
 
  • #13
haruspex
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Do you think 0.084 N is the correct answer?
Does adding 10% to 0.045 give 0.084?
 
  • #14
BurpHa
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Does adding 10% to 0.045 give 0.084?
No. But my book gave 0.084 N as the answer. I did get your result when I tried the problem.
 
  • #15
BurpHa
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Does adding 10% to 0.045 give 0.084?
There is a period when the acceleration increases, and that is what I don't understand. I mean, when you apply a force, the acceleration should be constant.
 
  • #16
haruspex
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There is a period when the acceleration increases, and that is what I don't understand. I mean, when you apply a force, the acceleration should be constant.
only if the force is constant.
 
  • #17
BurpHa
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only if the force is constant.
So you take a = 90 m / s ^ 2. But for the most part, the acceleration increases, so why you choose this acceleration? I did what you did until I realized the acceleration is not constant.
 
  • #18
Lnewqban
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There is a period when the acceleration increases, and that is what I don't understand. I mean, when you apply a force, the acceleration should be constant.
The acceleration (increasing velocity) of any constant mass is a consequence of the magnitude of the net force that is applied to it.

In this case, the net force increases as time goes by, reaches a peak, and decreases again.
The reason is the mechanism of the legs and muscles of the insect, which provide a different amount of force as they extend.

On the other hand, the weight (force induced by gravity) of the insect remains constant during the process.

The net force mentioned above is the vectorial summation of leg's force and weight.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html
 
  • #19
haruspex
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So you take a = 90 m / s ^ 2. But for the most part, the acceleration increases, so why you choose this acceleration?
Because it asks for the maximum force.
 
  • #20
BurpHa
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The acceleration (increasing velocity) of any constant mass is a consequence of the magnitude of the net force that is applied to it.

In this case, the net force increases as time goes by, reaches a peak, and decreases again.
The reason is the mechanism of the legs and muscles of the insect, which provide a different amount of force as they extend.

On the other hand, the weight (force induced by gravity) of the insect remains constant during the process.

The net force mentioned above is the vectorial summation of leg's force and weight.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html
Thanks, a bit of biology helps.
 
  • #21
BurpHa
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Because it asks for the maximum force.
I agree, but I don't know why my textbook gives the result as 0.084 N.
 
  • #22
kuruman
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I agree, but I don't know why my textbook gives the result as 0.084 N.
It could be a typo or it could be that the person who who provided the solution made a mistake. Wrong solution answers appear once in a while. I agree with your answer.
 
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  • #23
haruspex
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I searched for the question online. Got several hits, all the same graph, just small variations in locust mass.
All were on paywalled student help sites, though you could sign up for a free trial. The only one I looked further into was Brainly. The "expert verified" solution it showed quoted "Fnet=ma-mg". At that point I gave up.
 
  • #24
BurpHa
44
13
I searched for the question online. Got several hits, all the same graph, just small variations in locust mass.
All were on paywalled student help sites, though you could sign up for a free trial. The only one I looked further into was Brainly. The "expert verified" solution it showed quoted "Fnet=ma-mg". At that point I gave up.
Brainly is not reliable, those "expert verified" thing is rubbish. Yeah, most of the other sites are paid, that is why I look at my textbook's solution to check ;))) By the way, thank you for your help!!
 
  • #25
BurpHa
44
13
Homework Statement:: The jumping ability of the African desert locust was measured by placing the insect on a force plate, a platform that can
accurately measure the force that acts on it. When the locust
jumped straight up, its acceleration was measured to follow the
curve in Figure P4.71. What was the maximum force that this
0.50 g locust exerted on the force plate?
Relevant Equations:: Newton's Second Law: F = ma.

At the beginning, I just looked for the highest point in the graph, which is approximately 90 \frac m s^2.
Then I plugged it in the formula F = ma and got the force equaled to 0.045 N.

However, when I looked back, the graph is about the change in acceleration. So really, I'm dealing with acceleration of acceleration. From what I understand, the acceleration increases when the locust jumps up until t is approximately about 27 ms, when it starts to decrease.

The answer I get from my book is 0.084 N. When I plug 0.084 N to the formula F = ma, the acceleration is 168 \frac m s^2

I don't understand how it got 168 \frac m s^2 so that it could get 0.084 N.

Thank you for your help.
I have received a great deal of help! All I want to say is thank you!
 

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