MHB Challenging Classical Mechanics Problems: Can You Solve Them?

Click For Summary
The discussion focuses on solving two classical mechanics problems involving motion up and down an incline with kinetic friction. Key equations are provided for calculating distance and time, incorporating initial and final velocities, acceleration due to gravity, and friction. The second problem relates to d'Alembert's principle, emphasizing the use of fictitious forces in non-inertial frames to equate dynamics between different systems. The participant expresses gratitude for the assistance received, indicating that the help will aid in tackling similar problems. Overall, the thread highlights the challenges of applying classical mechanics concepts to specific scenarios.
johnherald
Messages
2
Reaction score
0
Hello i have the difficulty in solving this two problems..thank you for your help math help boards :-) View attachment 8748
 

Attachments

  • 5 & 6.png
    5 & 6.png
    15.5 KB · Views: 153
Mathematics news on Phys.org
1.6

distance up the incline (with kinetic friction present) ...

$\Delta x = \dfrac{v_f^2 - v_0^2}{2a}$

time up the incline ...

$t = \dfrac{v_f-v_0}{a}$

... where $v_f=0$ and $a = -g(\sin{\theta} + \mu \cos{\theta})$time down the incline ...

$\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2 \implies t = \sqrt{\dfrac{2\Delta x}{a}}$

note $v_0 = 0$, $\Delta x$ is the opposite of that found going up the incline and $a = -g(\sin{\theta} - \mu \cos{\theta})$
 
1.5 is related to d'Alembert's principle. We can analyze the dynamics of an accelerating frame of reference (i.e non-inertial) by adding a fictitious force. Since the force in system A is $m*100$, then a fictitious force of $m*10$ must be added to system B so that the dynamics in both systems are equivalent.
 
thanks a lot that will help me solving the other problems similar to that two particular question...:-)
and i have the difficulty solving some problems specially to the questions without a value.. thank you again for your help..
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
9
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 66 ·
3
Replies
66
Views
7K