Mass on a wedge. Relative acceleration.

Click For Summary
SUMMARY

The discussion focuses on the dynamics of a particle of mass M on a smooth wedge of mass 2M, inclined at 30 degrees. The problem involves calculating the acceleration of both the wedge and the particle relative to the wedge. Through the application of Newton's second law (F=ma) and trigonometric identities, the acceleration of the wedge is determined to be a = g√3/9, which aligns with the answer provided in the textbook, a = g/(3√3), after simplification.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of trigonometric functions, specifically sine and cosine
  • Familiarity with free body diagrams
  • Basic algebra for solving equations
NEXT STEPS
  • Study the application of Newton's laws in non-inertial reference frames
  • Learn about inclined plane dynamics and forces acting on objects
  • Explore advanced topics in particle dynamics, including frictionless surfaces
  • Review the derivation and application of trigonometric identities in physics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of inclined plane problems and their solutions.

Darth Frodo
Messages
211
Reaction score
1

Homework Statement



A particle of mass M is on a wedge of mass 2M. The wedge is smooth, and is inclined at 30* to the horizontal. The system is released from rest. Find the acceleration of the wedge, and find the acceleration of the particle relative to the wedge.

Homework Equations



F=ma

The Attempt at a Solution


Particles motion parallel to the wedge

F=ma
(mg)sin30 = m[f - (a)cos30]
mg = 2mf - ma\sqrt{3}Particles motion perpendicular to the wedge

F = ma
(mg)cos30 - R = m(asin30)
mg\sqrt{3} -2R = ma
R = \frac{m(g\sqrt{3} - a)}{2}Wedges Motion

F = ma
Rsin30 = 2ma
R = 4ma

Substitution

\frac{m(g\sqrt{3} - a)}{2} = 4ma

mg\sqrt{3} - ma 8ma

g\sqrt{3} - a = 8a

g\sqrt{3} = 9a

a =\frac{g\sqrt{3}}{9}

The answer at the back of the book is \frac{g}{3\sqrt{3}}
 
Physics news on Phys.org
Darth Frodo said:
a =\frac{g\sqrt{3}}{9}

The answer at the back of the book is \frac{g}{3\sqrt{3}}

Your solution is the same as that of the book. (Write 9 as 3*√3*√3 and simplify by √3.)

ehild
 

Similar threads

Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
11
Views
2K
Replies
7
Views
1K
  • · Replies 25 ·
Replies
25
Views
4K