Understanding the Chain Rule in Mechanics: Solving for Acceleration and Force

AI Thread Summary
The discussion focuses on applying the chain rule to find acceleration and force for a particle moving with a speed defined by v(x) = α / x. The user initially misunderstands the need for the chain rule, believing they can directly differentiate with respect to time. Clarification reveals that since x is a function of time, the chain rule must be used to correctly derive acceleration as a = (dv/dx)(dx/dt). The resulting force is confirmed to be F(x) = -m α^2 / x^3, aligning with the given answer. The thread concludes with a note about the forum's functionality regarding marking discussions as resolved.
AshesToFeonix
Messages
11
Reaction score
0

Homework Statement




6. A particle of mass m moves along a frictionless, horizontal plane with a speed given by

v(x) = α / x. Where x is the distance of the object from the origin and α is a constant.

Working with F = ma, we want to get the acceleration. You have v = v(x). You want a = dv/dt. Find (dv/dx)(dx/dt). Find the force F(x) to which the particle is subjected to.




The Attempt at a Solution



I guess my problem is I don't understand why I need to use chain rule since v = dx/dt. I thought I could take the derivative in respect to t on both sides, and get dv/dt = - α / x^2, then multiply both sides by m to get the force equation.

the answer is given, -m α^2/ x^3. So can someone explain what I'm missing here...
 
Physics news on Phys.org
You need to use the chain rule because x is some function of t. What you have done above is find dv/dx. Now you have correctly identified dx/dt as v and you know v = a/x, so what is (dv/dx)*(dx/dt)?
 
wow awesome thanks that clears up a lot. I almost gave up on anyone answering me. I read that there was a way to close a thread or say that the problem is solved but I'm not seeing it on here so I guess'll have to leave it as is.
 
The forum software was upgraded recently and I think only mentors can mark it solved at the minute. Just leave it as it is for now. :smile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top