Two Blocks and a Pulley Friction Problem

AI Thread Summary
The problem involves a system of three blocks connected by a pulley, with specific weights and friction coefficients. The 9.0 kg block is on a smooth table, while the 12 kg block experiences friction with a coefficient of 0.2. The calculations for the forces acting on each block were outlined, leading to the determination of tension and acceleration. The final acceleration of the 5.0 kg block, when released, was calculated to be 1.82 m/s². The solution appears to be confirmed as correct by the poster.
joemost12
Messages
20
Reaction score
0

Homework Statement



System comprised blocks, a light frictionless pulley and connecting ropes (see diagram). The 9.0kg block is on a perfectly smooth horizontal table. The surfaces of the 12kg block are rough, with μk = .2 between the two blocks. If the 5.0 kg block accelerates downward when it is released, find its acceleration

Fairly certain my solution is correct, just would love a second opinion!

Diagram:
6k5Im7W.png

http://i.imgur.com/6k5Im7W.png

Homework Equations

The Attempt at a Solution



So for the 12kg Block my reasoining for its forces was:
Fnet = Ffriction - T1 = Ma = 0
Fnormal = (12kg)(9.8 m/s^2 _ = 117.6
ForceFriction = (.2)(11.6) = 23.52 N

Then for the 9 KG block:
Fnet = T2 - ForceFriction
T2 - 23.52N = Ma
T2 - 23.52N = (9kg)a
23.52-T2 = (9 kg) a

Then finally for the hanging 5.0 KG block:
T2 - Fg = ma
T2 - (5)(9.8) = (5)a
T2 - 49N = 5(a)

Then I simply solved the system of equations to find the value of A

Set two equations for the 9kg and the 5kg block in terms of a

(T2 - 49N)/5 kg = (23.52N - T2)/9kg
Solving to find that T2 = 39.9
Then simply plug that value back into T2 - 49N = 5a

a = 1.82 m/s^2
 
Physics news on Phys.org
Looks good.
 
gneill said:
Looks good.
Great. Thanks
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top