Force & Acceleration: Explaining F=ma in Disguise

  • Thread starter Thread starter 404
  • Start date Start date
  • Tags Tags
    Stupid
AI Thread Summary
The discussion centers on explaining the equation F=ma using a force vs. acceleration graph, where force is represented on the y-axis and acceleration on the x-axis, with mass as the slope. Participants note that while the teacher requested not to use traditional x and y labels, the relationships between the variables remain crucial. The linear relationship indicates that force is directly proportional to acceleration, scaled by mass. The conversation emphasizes understanding the graph's structure and the significance of the slope in revealing the mass. Ultimately, clarity in expressing these relationships without standard axis labels is the key takeaway.
404
Messages
52
Reaction score
0
Ok we did a force vs acceleration graph in class, and got an equation from it(couldnt use Y and X for the equation). The question is explain why the equation is F=ma in disguise. Obviously it is, like force is the y, slope is the mass etc, but how can I actually explain it is?
 
Physics news on Phys.org
So force is your y, acceleration is x, and mass is the slope. Do you know the relationship found in the graph that relates to the axes and slope?
 
For a linear equation y=mx+b...

Here we are neglecting b, the y-intercept. What does that leave us?
 
But the teacher said we couldn't use X and Y...
 
I don't understand what he/she wants from you. How are you supposed to explain the relationship between variables on a linear graph without using x and y??

The only way I can see how is to generalize and say that basically a linear graph has these relationships with the axes and slope... but without mentioning the axes that will be a bit vague.
 
The acceleration is the indep. variable, the force is the dep variable. They have a directly proportional relationship by a scalar quantity 'm'
 
may be what your teacher wants is to show that
Y=F (force)
X=a (acceleration)
m(gradient)=m(mass)
 
Alight I got it now, thanks.
 
Back
Top