Question

ess: U10 Our payme Please ne QUESTION 5 elp avall 150 g a) A 120g trolley on a 450 inclined plane and a 150 g trolley on a 300 inclined plane are attached by a light string over a pulley, as shown in the diagram. Friction force can be ignored. Determine the acceleration of the system and the tension in the string? ave troub dential cheme provid 120 g mo 45° F = ma 300 more

Answer 1

T a T M Mgsin45° mgsin30° 30° 45°

Assume that the acceleration is along the 45° incline.

We have:

M = 150 g = 0.150 kg

m = 120 g = 0.120 kg

g = 9.8 m/s²

Sin 30° = 0.5

Sin 45° = 0.7071

For the left incline:

T - mgsin30° = ma --------- (1)

For the right incline:

Mgsin45° - T = Ma --------- (2)

(1) + (2) ====>

Mgsin45°- mgsin30° = (M+m)a

a = (Mgsin45°- mgsin30°) / (M+m)

a = (0.150*9.8*0.7071 - 0.120*9.8*0.5) / (0.150 + 0.120)

a = (0.451) / (0.270)

a = 1.672 m/s²------------- (**Answer**)

(since acceleration is positive, the direction of acceleration is along the 45° incline. So our assumption was true)

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From equation (1):

T - mgsin30° = ma

T = mgsin30° + ma

Substitute, a = 1.672 m/s²

T = (0.120*9.8*0.5) + (0.120*1.672)

T = 0.7886 N ---------- (**Answer**)

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