- #1
Woolyabyss
- 143
- 1
Homework Statement
Two particles of masses 6 kg and 8 kg rest on two horizontal tables.The coefficient of friction between both particles and their respective tables is (1/2).The particles are connected by a smooth inextensible string which passes over smooth pulleys and under a smooth movable pulley of mass M kg
Show that none of the particles will move if M is less than or equal to 6
Homework Equations
F=MA
The Attempt at a Solution
t = Tension
a = acceleration of 6 kg particle
b = acceleration of 8 kg particle
for the six kg particle 3g - t = 6a
a = (3g - t)/6
for the eight kg particle 4g - t = 8b
b = (4g-t)/8
for pulley M
2t -Mg = M(a+b)/2
4t -2Mg = M(3g-t)/6 + M(4g-t)/8
96t -48Mg = 4M(3g-t) + 3M(4g -t) multiplied my 24...
96t = 48Mg + 12Mg -4Mt + 12Mg - 3Mt
96t + 7Mt = 72Mg
t(96 + 7M)= 72Mg
t = 72Mg / (96 + 7M)
Since its not accelerating
2t - Mg = (3g - t) + (4g -t) Sum of the forces are equal
4t= Mg + 7g
288Mg = (Mg + 7g)(96 + 7M) Substituting value for t
288Mg = 96Mg + 672g +7Mg^2 + 49Mg
7Mg^2 - 143Mg +672g = 0
68.6M^2 - 1401.4M + 6585.6
When I use the minus b formula i get M = 13.1 and 7.32
I'm convinced I've done this completely wrong.I was hoping I would get a solution of M = 6.
Any help would be appreciated.