Three blocks of masses m1 = 2 kg, m2 = 4 kg, and m3 = 6 kg are connected by strings on an
inclined plane of 60 as shown. A force of Fapplied = 120 N is applied upwards along the incline
to the uppermost block, causing an upward movement of the blocks along the incline. The
connecting strings are massless.
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Answered by
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Explanation:
Fapplied = 120 N is applied.....
Answered by
1
Answer:
given = m1 = 2kg
m2 = 4kg
m3 = 6kg
θ = 60°
f = 120N
to find = accelaration
solution =
use the newtons second law of ,motion
F - T2 -Mgsin60° =m3a ----- 1
T2 - T1 -mgsin60° = m2a-----2
T1-T1- mgsinsiii60° = m1a -----3
add all above 3 equations
F- (m1+m2+m3) gsin60° = (m1 +m2 +m3 ) a
120-(2+4+6) 9.8 X √3/2 = (2+4+6)a
a= 1.51 m/s²
the accelaration is a= 1.51 m/s²
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