# Homework Help: Acceleration of wedges and particles

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1. Jun 13, 2013

### Woolyabyss

1. The problem statement, all variables and given/known data
A wedge, all of whose faces are smooth, resting on a smooth horizontal table.Its faces make an angle A and B with the table, where tan A = 4/3 and tan b =3/4 .Particles of each of mass m are placed on these faces.If the mass of the wedge is 2m, show that the wedge will not move

2. Relevant equations

F = MA

3. The attempt at a solution
I tried to make equations for both particles seperately

first particle
tan A = 4/3
forces and acceleration of particle resolved in to components

parallel = mg(4/5) perpendicular = mg(3/5)
parallel = a(3/5) perpendicular = a(4/5)
a = acceleration of wedge
f= acceleration of particle relative to wedge
R1=normal force

equations
(4/5)mg = m(f - (3a/5) )................. simplify 4g = 5f - 3a (I didn't use this equation)

3/5)mg - R1 = 4ma/5 ......................... simplify 3mg - 5R1 = 4ma

Forces and acceleration of the wedge along horizontal
R1(sinA)........... (4/5)R1 = 2ma simplify R1 = (10/4)ma

putting value for R1 into second equation 3mg - (50/4)ma = 4ma

simplify a = (2/11)g

Second particle
tan B = 3/4

forces and acceleration of particle resolved in to components
parallel = (3/5)mg perpendicular = (4/5)mg
parallel = (4/5)b perpendicular = (3/5)b

b = acceleration of wedge
e = acceleration of particle relative to wedge
R2 =normal force

equations

(3/5)mg = m(e - 4b/5 ) (I didnt use this equation)

(4/5)mg - R2 = (3/5)b simplify 4mg - 5R2 = 3mb

Acceleration of wedge along the horizontal

R2(sinB) R2(3/5) = 2mb ......... R2= (10/3)mb

put R2 into second equation

4mg - (50/3)mb = 3mb .....multiply by 3.... 12mg = 50mb + 9mb

59b = 12g b = (12/59)g

a-b = (2/11)g - (12/59)g is not zero

I had no problem with these questions when there was only one particle on the wedge.Now that there is two particles im not sure how to solve them. Should I have tried to make equations including both the particles instead of trying to solve both seperately? Any help would be appreciated.

2. Jun 13, 2013

### haruspex

Since you are only required to show that the wedge remains still, rather than calculate an acceleration, I feel it would be simpler to suppose another force F is applied to the wedge to hold it still then calculate F (and show it is 0). That avoids getting tangled up in relative accelerations.

3. Jun 14, 2013

### Woolyabyss

Sorry im not exactly sure how I can only take into account the forces without the acceleration?

4. Jun 14, 2013

### haruspex

I'm saying:
- suppose there is also a horizontal force F on the wedge sufficient to keep it still
- now you can assume the wedge is stationary, and figure out the forces on and accelerations of the two particles in the usual manner
- show that F=0

5. Jun 14, 2013

### Woolyabyss

Would this be correct?

r1 and r2 are the normal forces of the particles

r1 = mg(cosa) = 3mg/5 and r2 = mg(cosb) =4mg/5

the forces acting on the wedge horizontally are F1 and F2

r1(sina) = 4r1/5 = (4/5)(3mg)/5) = 12mg/25 = F1

and the force in the opposite direction

r2(sinb) = 3r2/5 =(3/5)(4mg/5) = 12mg/25 = F2

Subtracting F1 and F2..... 12mg/25 - 12mg/25 =0 N

Since the net force acting on the wedge horizontally equates to zero the wedge does not move.

6. Jun 15, 2013

### haruspex

That works for me.

7. Jun 15, 2013

### Woolyabyss

Alright thanks for the help.

8. Jun 15, 2013

### Woolyabyss

Sorry again but I just realized I might have made a mistake. Should
R1 = 3/5)mg - 4ma/5

and

R2 = (4/5)mg - (3/5)b

????

9. Jun 15, 2013

### haruspex

You considered forces normal to the surface, right?
In the normal direction, there is no acceleration of the particle.

10. Jun 16, 2013

### Woolyabyss

I got somebody to help me do it the way I originally tried(with the accelerations).
But thanks anyway