Graphical Impulse calculation from the graph of force verus time

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Homework Help Overview

The discussion revolves around calculating impulse from a force versus time graph. The original poster attempts to use both area calculations and a formula involving force and time, but encounters difficulties due to the non-constant nature of the force and uncertainty about the shape of the area under the graph.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore different methods for calculating impulse, including area under the curve and applying the impulse formula. Questions arise about the calculations made and the interpretation of the graph's shape.

Discussion Status

Participants are actively questioning the original poster's calculations and reasoning. Some guidance has been offered regarding breaking down the area into simpler shapes and considering the implications of non-constant force.

Contextual Notes

There is mention of specific time intervals and force values, but the original poster expresses confusion about the shape of the area to be calculated, indicating a potential gap in understanding the graphical representation.

yesmale4
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Homework Statement
Shown below is a graph of the x-component of the force as a function of time FYP, x(t) that you exert on a particle of mass 18 kg.
FYP, x(t) = 0 at t = 18.095 s.
The particle starts at t = 10 s with an initial x-component of its velocity of 6.420 m/s and moves under the influence of only the force shown in the graph.

What is the x-component of the impulse that you give to the particle from t = 10.00 s until t = 33.00 s.
Relevant Equations
J=F*(t-t0)
area of triangle
area of rectangle
d.png


hello i would like to get some help with this problem.
At first it try to calculate the impulse by the area but i found it too difficult
Then i try to solve it by the formula J= F(t-t0), but the problem is that i don't know what F is so i try to solve it like this
F10) = 8
F(33) = -13
so
EF = 8+-13 = -5
T = 33 -10 = 23
F = 23 * -5 = -115
and it wrong so i would like to know what I am supposed to do here
 
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yesmale4 said:
EF = 8+-13 = -5
What do you think you have calculated there?
How did you try to do it by graph area? Describe the steps.
 
haruspex said:
What do you think you have calculated there?
How did you try to do it by graph area? Describe the steps.
about the first question i was thinking that because impulse formula is J=F*(t-t0) and the force isn't constant then if i calculate the change in the force and multiply it by the change of time it will be alright
about the second question i was not able to calculate the area between t = 12 to t= 28 because i don't undertand what kind of shpe it is , its not triangle so that's why i try diffrent method
 
yesmale4 said:
i was thinking that because impulse formula is J=F*(t-t0) and the force isn't constant then if i calculate the change in the force and multiply it by the change of time it will be alright
If a car drove at one speed for a while, slowed down to a different speed, then drove at that speed for a while, would you find the distance traveled by multiplying the change in speed by the time taken?
What if you were to increase the speed by 1m/s at all stages? You would calculate the same number. What if you increased the time at the higher speed and reduced the time at the lower speed by the same? Again, you would calculate the same number.
yesmale4 said:
i was not able to calculate the area between t = 12 to t= 28 because i don't undertand what kind of shpe it is , its not triangle
So cut it into simpler shapes.
 
From t=10 s to t = 12 s the force is constant: F = 8 N
Can you do that part of the exercise ?

And: please use units

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