What is the y-component of the force by the floor on her during the collision?

In summary, the y-component of the force exerted by the floor on an object during a collision is determined by analyzing the forces acting on the object, particularly its weight and any acceleration it may experience. This involves applying Newton's second law of motion, where the net force in the vertical direction equals the mass of the object multiplied by its acceleration. If the object is in free fall, the force will counteract its weight, while during impact, the force may exceed the weight temporarily due to the deceleration.
  • #1
ryancerealbox
8
0
Homework Statement
Your lab partner jumps off a lab table and lands on the floor. Using video analysis, you measure her y-velocity during a certain time interval during the collision with the floor and graph 𝑝𝑦(𝑡) as shown below. Her mass is 63 kg. The slope of the best-fit line is 2625 kg·m/s2.

https://education.wiley.com/prod/w660d7fcb-fe26-43e3-9f17-93c3fbe62310/art/qb/qu/ch0/EAT_1593581387456_1_8615938273917383/qu/ch0/EAT_15932383140560_8051448095064999.png

2. What is the y-component of the force by the floor on her during the collision?

3. How does the force by the floor on her during the collision compare to the force by Earth?
Relevant Equations
|F floor| / |F gravity|
A previous question I got correct was: During this time interval, what is the y-component of the average net force on her? It was just 2625, and I thought question 2 would be the same answer but it wasn't. Not sure what to do.
 
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  • #2
I'm a bit confused about what's being measured by what here? ##kg\cdot m/s^2## is newtons, but how do you measure force by video analysis? Also, the floor is the Earth for all practical purposes.
 
  • #3
Sorry, forgot to include the actual graph.
1706298365676.png
 
  • #4
ryancerealbox said:
[...]
2. What is the y-component of the force by the floor on her during the collision?
[...]
During this time interval, what is the y-component of the average net force on her? It was just 2625, and I thought question 2 would be the same answer but it wasn't. Not sure what to do.
One question asks for the net. The other question asks for one of the contributions to the net. Can you list the forces that act on her during the collision?
 
  • #5
jbriggs444 said:
One question asks for the net. The other question asks for one of the contributions to the net. Can you list the forces that act on her during the collision?
The floor's opposite reaction and gravity? So 2625 and -9.8?
 
  • #6
ryancerealbox said:
The floor's opposite reaction and gravity? So 2625 and -9.8?
Be careful. One of those numbers is a force. The other is an acceleration.

Being careful in this case might include putting units after your numbers. That could help keep you on the straight and narrow.

Note that 2625 [Newtons] is the net force, not the normal force.
 
  • #7
jbriggs444 said:
Be careful. One of those numbers is a force. The other is an acceleration. But yes, you have identified the two relevant forces.

Being careful in this case might include putting units after your numbers. That could help keep you on the straight and narrow.
Since gravity is measured in m/s^2 its acceleration then, so the only force would be the 2625 kg*m/s^2? When I put that in it was wrong though
 
  • #8
ryancerealbox said:
Since gravity is measured in m/s^2 its acceleration then, so the only force would be the 2625 kg*m/s^2? When I put that in it was wrong though
No, you had the right idea in post #5 but you need two forces, not a force and an acceleration.
 
  • #9
So I multiply -9.8 by the mass 63 to get 617.4 kg*m/s^2 then subtract that from 2625 to get 2007.6 as the total force
 
  • #10
ryancerealbox said:
So I multiply -9.8 by the mass 63 to get 617.4 kg*m/s^2 then subtract that from 2625 to get 2007.6 as the total force
I thought 2625 was the total force. Q2 asks for the force from the floor only.
 
  • #11
ryancerealbox said:
So I multiply -9.8 by the mass 63 to get 617.4 kg*m/s^2 then subtract that from 2625 to get 2007.6 as the total force
If you manipulate the numbers enough times, you must get the right answer eventually!
 
  • #12
haruspex said:
I thought 2625 was the total force. Q2 asks for the force from the floor only.
My bad, so 2007.6 would be just the force from the floor, but that isn't the answer either.
 
  • #13
ryancerealbox said:
My bad, so 2007.6 would be just the force from the floor, but that isn't the answer either.
Write the equation relating the two forces to the net force.
 
  • #14
haruspex said:
Write the equation relating the two forces to the net force.
2625 kg*m/s^2 = 2007.6 kg*m/s^2 + 617.4 kg*m/s^2
 
  • #15
Isn't force a vector?
 
  • #16
I forgot about the negative, thanks guys it was 3242
 

FAQ: What is the y-component of the force by the floor on her during the collision?

What is the y-component of the force by the floor on her during the collision?

The y-component of the force by the floor on her during the collision is the vertical force exerted by the floor, which counteracts her weight and any additional vertical forces due to the collision.

How do you calculate the y-component of the force by the floor on her during the collision?

To calculate the y-component of the force by the floor, you need to know her mass, the acceleration due to gravity, and any additional vertical acceleration caused by the collision. The force can be found using Newton's second law, F = ma, where 'a' is the total vertical acceleration.

Is the y-component of the force by the floor on her always equal to her weight?

No, the y-component of the force by the floor on her is not always equal to her weight. During a collision, additional vertical forces may act on her, causing the y-component of the force to be greater or less than her weight.

How does the duration of the collision affect the y-component of the force by the floor?

The duration of the collision affects the y-component of the force by the floor because a shorter collision time generally results in a larger force due to the higher acceleration needed to change her momentum in a shorter period.

Can the y-component of the force by the floor be negative during a collision?

The y-component of the force by the floor can be considered negative if it acts in the opposite direction to the positive y-axis, typically downward. However, in most scenarios involving a collision with the floor, this component is positive as the floor pushes upward against her.

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