(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

John (mass = 80 kg) rests on the top face of a light cube (mass = 0). A light rope passes horizontally from him over the front edge, and vertically down to Brian (mass = 60 kg) who hangs in contact with the front face as shown. The cube rests on a horizontal surface and there is no frictional force anywhere. Joe (mass = 70 kg) has his back against a wall and applies a horizontal push perpendicular to the rear face. Calculate the value of this force, which will prevent both John and Brian from sliding over the surface of the cube.

2. Relevant equations

F_{UN}= ma

Fg = mg

3. The attempt at a solution

John (80 kg) = m_{1}

Brian (60 kg) = m_{2}

Fg_{2}= m_{2}g

Fg_{2}= (60)(9.8)

Fg_{2}= 588 N

Since John and Brian should not slide off the cube, I'm guessing the force of tension in the rope should have the same magnitude as the force of gravity.

FT = Fg_{2}

FT = 588 N

F_{UN}= m_{1}a_{1}

FT = m_{1}a_{1}

a_{1}= FT/ m_{1}

a_{1}= 588/ 80

a_{1}= 7.35 m/s²

The light box will need a force to accelerate it at a rate of 7.35m/s².

The mass of the box will be equal to 60 kg because it is in contact with Brian.

F_{UN}= m_{3}a_{1}

F_{A}- F_{On 3 due to 2}= m_{3}a

F_{A}= m_{3}a + F_{On 3 due to 2}

F_{A}= (0)(7.35) + (m_{2}a_{2})

F_{A}= 0 + (60)(7.35)

F_{A}= 441 N

I'm not sure where I went wrong, but it doesn't seem right. Thank you very much for any help.

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# Homework Help: Newton's Laws, Finding Correct Acceleration and Applied Force

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