Newton's Laws, Finding Correct Acceleration and Applied Force

[Error]

Homework Statement

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.

Homework Equations

F_UN = ma
Fg = mg

The Attempt at a Solution

John (80 kg) = m₁
Brian (60 kg) = m₂

Fg₂ = m₂g
Fg₂ = (60)(9.8)
Fg₂ = 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 = Fg₂
FT = 588 N


F_UN = m₁a₁
FT = m₁a₁
a₁ = FT/ m₁
a₁ = 588/ 80
a₁ = 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 = m₃a₁
F_A - F_{On 3 due to 2} = m₃a
F_A = m₃a + F_{On 3 due to 2}
F_A = (0)(7.35) + (m₂a₂)
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.

 

[gneill]

Homework Statement

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.

Homework Equations

F_UN = ma
Fg = mg

The Attempt at a Solution

John (80 kg) = m₁
Brian (60 kg) = m₂

Fg₂ = m₂g
Fg₂ = (60)(9.8)
Fg₂ = 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 = Fg₂
FT = 588 N


F_UN = m₁a₁
FT = m₁a₁
a₁ = FT/ m₁
a₁ = 588/ 80
a₁ = 7.35 m/s²


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

Excellent up to here!

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

Joe needs to accelerate the whole system, including both John and Brian. In fact, it's Brian's inertial force that counteracts the tension in the rope he's holding.

 

[Error]

Oh! I think I understand! Instead of putting a mass of 0, I should use 80 kg because that is John's mass?

I would get:

F_UN = m₃a₁ - m₂a₁
F_A = (80)(7.35) - (60)(7.35)
F_A = 147 N

Is 147 N correct?

 

[gneill]

Oh! I think I understand! Instead of putting a mass of 0, I should use 80 kg because that is John's mass?

I would get:

F_UN = m₃a₁ - m₂a₁
F_A = (80)(7.35) - (60)(7.35)
F_A = 147 N

Is 147 N correct?

No, Joe needs to produce this acceleration while pushing BOTH Brian and John. The masses add.

 

[Error]

If I add John and Brian's mass together, I think I should get:

F_UN = (m₁ + m₂)a₁
F_A = (80 + 60)(7.35)
F_A = 1029 N

 

[gneill]

If I add John and Brian's mass together, I think I should get:

F_UN = (m₁ + m₂)a₁
F_A = (80 + 60)(7.35)
F_A = 1029 N

That looks good.

 

[Error]

Thank you very much!