- #1
Woolyabyss
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Homework Statement
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
Homework Equations
F = MA
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 into 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 into 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 I am 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.