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I calculated the timeView attachment 11166

Friction= 1/(sqroot12)×80 cos 30= 20N

I get two equations

T-20=8a and 120-T=12a

a=5m/s^2

I don't know how to calculate the time. Also with this value of acceleration tension value is wrong and the textbook ans is 84N

S=ut +1/2at^2

(2+1.5)= 1/2×5×t^2

I got t=1s

For q(b) iam getting the ans tension in the rope= 60N but the textbook ans is 84N. Pls help

- Mar 1, 2012

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$T - mg(\sin{\theta} + \mu \cos{\theta}) = ma$

———————————————————————

$Mg - mg(\sin{\theta} + \mu \cos{\theta}) = a(M+m)$

$a = \dfrac{g[M-m(\sin{\theta}+ \mu \cos{\theta})]}{M+m} = \dfrac{10\left[12-8\left(\frac{3}{4}\right) \right]}{20}= 3 \, m/s^2$

$T = M(g-a) = 12(7) = 84$N

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Thank you very much!!!

$T - mg(\sin{\theta} + \mu \cos{\theta}) = ma$

———————————————————————

$Mg - mg(\sin{\theta} + \mu \cos{\theta}) = a(M+m)$

$a = \dfrac{g[M-m(\sin{\theta}+ \mu \cos{\theta})]}{M+m} = \dfrac{10\left[12-8\left(\frac{3}{4}\right) \right]}{20}= 3 \, m/s^2$

$T = M(g-a) = 12(7) = 84$N