Can i get help with my F=ma problems?

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SUMMARY

This discussion focuses on solving physics problems using Newton's second law, F=ma, and emphasizes the importance of free body diagrams. The participants worked through four specific problems involving forces in an elevator, maximum occupancy calculations, and friction on a lawnmower. Key insights include the necessity of identifying all forces acting on an object and the correct application of trigonometric functions to resolve forces. The discussion highlights that drawing diagrams and listing known variables are essential steps in solving physics problems effectively.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Ability to draw free body diagrams
  • Basic knowledge of trigonometric functions (sine, cosine, tangent)
  • Familiarity with units of force (Newtons) and mass (kilograms)
NEXT STEPS
  • Learn how to draw and interpret free body diagrams in physics problems
  • Study the application of trigonometric functions in resolving forces
  • Explore advanced concepts in dynamics, such as tension and friction coefficients
  • Practice solving real-world physics problems involving elevators and friction
USEFUL FOR

Students studying physics, particularly those struggling with dynamics and force analysis, as well as educators seeking effective teaching strategies for explaining Newton's laws.

  • #31
Netgypsy - 650N/9.8m/s^2 gives the mass of the person on the balance.
 
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  • #32
duhhh right but I thought they were on an elevator?
 
  • #33
On a balance on an elevator!
 
  • #34
OK I hate it that I can't see the problem cause I'm on the second page. I should open the first in a different window but I'm too brain dead.
 
  • #35
Now to number 3. I would do tan(28)=129/fx correct? Then fx-68 to find the answer?
 
  • #36
Pretty sure you don't want to use tan in this situation. Drawn a diagram?
 
  • #37
Again draw the forces ON the lawnmower. You have the weight which acts down. The push on the handle which acts diagonally DOWN from the middle of the mower and there is an upward force of the ground on the mower. You draw the friction parallel to the ground and in the backward direction.

Since the mower is moving parallel to level ground you are only concerned with forces that act parallel to the ground. How many do you have? Well obviously friction but is there another one? If so what is it?
 
  • #38
So sin? because this is where i am confused. I don't know if i want to find the hypotinus or the just left to right force
 
  • #39
Only forces that are parallel to the ground aka left and right

sin will give you a downward force So what's left to use?
 
  • #40
You want the total force applied by the student, so that'd have to be the hypotenuse. But you'll have to find the parallel forces first.
 
  • #41
okay. So i want to find the force that is parrelel to the ground (other then friction). So to use one of the sin, cos, tan i need another side to the triangle. Would it be the 14.9kg that is pushing the lawnmower down?
 
  • #42
In other words you can find the parallel force first, then use the trig function to find the diagonal if you wish or you can do it all in one equation.

The sum of the forces in the x direction equal mass times acceleration. One force moves the mower, the other holds it back.

Nope - the weight is vertical. You don't need it in this problem to figure out the horizontal motion since friction is already given

What's the only thing you don't know?
 
  • #43
You can use the letter F for a force you don't know.
 
  • #44
If the sum of the forces in the x direction equal mass time acceleration, then that would mean 12.9*3.7=47.73. But that's less then the friction.. so wouldn't that mean that it is not moving?
 
  • #45
Think of it this way. When you push a lawn mower your push does two things. It actually makes the mower feel heavier because part of your push is downward. It makes the mower accelerate because it moves it forward
 
  • #46
OK your equation will be Force of the push in the x direction minus force of friction equals mass times acceleration
 
  • #47
You know all of this but the force of the push in the x direction so just put a letter in for that force. Like Fx
 
  • #48
Once you find that you can use your trig functions to find the diagonal force (hypotenuse)
 
  • #49
Fx-Ff=ma... So i know Fx-68=12.9*3.7
 
  • #50
right! Now just solve for Fx
 
  • #51
115.73 Now is this when i plug into cos? cos(28)=115.73/x (with x=hypotinus)
 
  • #52
You got it. Check your math. Good JOB
 
  • #53
Sick! i can't thank you enough. I would of got none of the equations right if it weren't for this. Website crashed again (probably because of all the kids doing homework last minute) but ill get it in as soon as it is up! thanks again and good night.
 
  • #54
Did you get number 2? It works like number 1 but don't forget to calculate the total mass and the total actual weight before you sum your forces.

Good luck
 
  • #55
Yah i got number 2. I was able to figure it out all by myself :D. Drawing things out makes it sooooo much easier.
 
  • #56
Tell me about it. I flunked my first physics courss (took it over and got an A though)
 
  • #57
i did the 70kg times the 20 (because it was the maxium occupancy). and then added the answer to that to 540kg to get the total down force. I converted it to Newtons then subtracted the up force (32471) from the down force. With that, I did F=Ma and got the answer. hopefully that made sense haha
 
  • #58
Drawing diagrams is a good rule of thumb in every physics course. I'm glad you got it all in the end.
 
  • #59
Yah and it helped that you guys both walked me through the steps rather then just doing it for me
 
  • #60
sounds good (sometimes two cooks make a better meal) hahaha
 

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