Question
Asked Oct 2, 2019
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As shown below, if F = 60.0 N and M = 4.00 kg,
what is the magnitude of the acceleration of the suspended
object? All surfaces are frictionless.

 

 

30°
2M
M
LL
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30° 2M M LL

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Expert Answer

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Step 1

The force acting on 2M can be resolved into two, Fsin(theta), and Fcos(theta).

Equation of motion of M and 2M can be written as.

 

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For mass M Mg-T Ma For mass 2M T -F cos30 2Ma NFsin 30° 2Mg 0 (II) (Ш)

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Step 2

Add equation (I) and (II), and obtain ...

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Mg F cos30 = 3Ma [M 4.00kg Mg -Fcos30 a 60.0 N g 9.8m/s Зм F = 60.0N (4.00kg) 9.8m/s2)-(60.0N) cos30 3(4.00kg) =-1.06m/s2

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