Welcome to our community

Be a part of something great, join today!

Physics Mechanics- connected particles

Shah 72

Member
Apr 14, 2021
193
20210530_210957.jpg
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
 

Shah 72

Member
Apr 14, 2021
193
View 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
I calculated the time
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
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
$Mg - T = Ma$
$T - mg(\sin{\theta} + \mu \cos{\theta}) = ma$
———————————————————————
$Mg - mg(\sin{\theta} + \mu \cos{\theta}) = a(M+m)$

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

$T = M(g-a) = 12(7) = 84$N
 

Shah 72

Member
Apr 14, 2021
193
$Mg - T = Ma$
$T - mg(\sin{\theta} + \mu \cos{\theta}) = ma$
———————————————————————
$Mg - mg(\sin{\theta} + \mu \cos{\theta}) = a(M+m)$

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

$T = M(g-a) = 12(7) = 84$N
Thank you very much!!!