Three blocks of masses 2 kg, 4 kg and 6 kg arranged as shown in figure connected by string on a frictionless incline of 37°. A force of 120 N is applied upward along the incline to the uppermost block. The cords are light. The tension T1 and T2 in the strings are (g = 10 m/s2]
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(A) T1 = 8N (B) T2 = 36 N (C) T1 = 20 N (D) T2 = 60 N
consider whole block as a system
then f =ma
or a = f/M M = m 1+m 2+m 3
a = 120/ 2+4+6 = 10 m/s 2
this is the common acceleration of each block
then
we have F- t 2 = 6*10
or t 2= 120-60 = 60 N
and
t 2- t 1 = 4*10
t 1= 20 N
then f =ma
or a = f/M M = m 1+m 2+m 3
a = 120/ 2+4+6 = 10 m/s 2
this is the common acceleration of each block
then
we have F- t 2 = 6*10
or t 2= 120-60 = 60 N
and
t 2- t 1 = 4*10
t 1= 20 N