Acceleration and tension for two blocks connected to frictionless pulley

[lil2ishaq]

I have tried to get this problem many different ways but still don't get it.

A 1.00 kg aluminum block and a 8.00 kg copper block are connected by a light string over a frictionless pulley. The two blocks are allowed to move on a fixed steel block wedge (of angle θ = 40.0°) as shown in Figure P4.63.

View attachment p4-63.bmp
Figure P4.63

Making use of Table 4.2, determine the following.
(a) the acceleration of the two blocks

Your answer differs from the correct answer by 10% to 100%. m/s2
(b) the tension in the string N


Coefficients of Frictiona µs µk
Steel on steel 0.74 0.57
Aluminum on steel 0.61 0.47
Copper on steel 0.53 0.36
Rubber on concrete 1.0 0.8
Wood on wood 0.25-0.5 0.2
Glass on glass 0.94 0.4
Waxed wood on wet snow 0.14 0.1
Waxed wood on dry snow - 0.04
Metal on metal (lubricated) 0.15 0.06
Ice on ice 0.1 0.03
Teflon on Teflon 0.04 0.04
Synovial joints in humans 0.01 0.003
a All values are approximate.

 

[asura]

I need help with this too... I keep getting a negative acceleration and I'm pretty sure I did everything correctly.

Cu block = 8 kg
Al block = 1 kg
theta = 30
∑Fx (Al) = ma = T - Fs ---> T = Fs + ma
∑Fx (Cu) = ma = sin 30 x mg - T - Fs

Substitution yields...
a (m + m) = sin 30 x mg - Fs (Al) - Fs (Cu)
(m + m ) = mass of Cu + mass of Al

Fs (Cu) = cos 30.0 x mg x 0.53 = 36.0 N
Fs (Al) = mg x 0.61 = 5.98 N

when I solve for a I get -0.31 m/s^2, which is impossible
so I thought a would be zero, but according to the website that's also wrong

 

[lil2ishaq]

anyone else that can help us with this

 

[lil2ishaq]

I tried to do this problem like how you did, but ended up in the same place.