Find force so that the mass does not move

  • Thread starter Calpalned
  • Start date
  • Tags Tags
    Force Mass
2 replies · 4K views · last reply
Calpalned
Messages
297
Reaction score
6

Homework Statement


Diagram of three blocks: block mA sits on top of large block mC, connected by a cord over a pulley at the upper right to hanging block mB; a horizontal force F pushes mC from the left, on a frictionless floor.
[/B]
Determine a formula for the magnitude of the force F exerted on the large block (mC) in the figure so that the mass mA does not move relative to mC. Ignore all friction. Assume mB does not make contact with mC.

Homework Equations


F = ma

The Attempt at a Solution



Whiteboard handwritten derivation: three boxes A, B and C with forces F and masses MBg and MA+MB, ending with the word WRONG in red; a smiley face is drawn beside 'Calpalned'.

Thank you all so much.
 
The blocks stay fixed in their relative positions. So to accelerate the whole assemblage: F = (mA + mB + mC).a

Does the textbook provide the correct answer? I'm not sure whether it's accurate enough to say the tension in the rope simply overcomes B's weight; you may need to acknowledge that the tension also provides B's horizontal acceleration. (Hence the rope's deviation from the vertical.)
 
Last edited:
Calpalned said:

Homework Statement


Diagram of a frictionless pulley system: block mA sits on top of larger block mC, which has a horizontal force F applied to its left face; a string runs over a pulley at the upper right and hangs down to block mB.
[/B]
Determine a formula for the magnitude of the force F exerted on the large block (mC) in the figure so that the mass mA does not move relative to mC. Ignore all friction. Assume mB does not make contact with mC.

Homework Equations


F = ma

The Attempt at a Solution



Handwritten whiteboard derivation: a sum of horizontal forces on B equals MBg equals (MA+MB)a, so aA = aB = MBg/(MA+MB). Below, a claim that aC = aA gives sum of Fx on C = McA, which becomes Mc(MBg/(MA+MB)), marked WRONG in red.

Thank you all so much.
You almost had it right, but not quite. The key to doing this problem correctly is to get the force balance on mass B correct.

Let T be the tension in the string and let θ represent the angle that the string attached at mass B makes with the vertical (i.e., the string is not vertical). In terms of T and θ, what is the horizontal component of the string tension on mass B? In terms of T and θ, what is the vertical component of the string tension on mass B? How does the horizontal component of acceleration of mass B compare with the acceleration a of mass A? What is the vertical component of the acceleration of mass B? In terms of T, θ, mB, and a, what is the horizontal force balance on mass B? In terms of T, θ, mB, what is the vertical force balance on mass B. In terms of T, mA, and a, what is the horizontal force balance on mass A?

If you did all this correctly, you will now have three equations in three unknowns: T, θ, and a. Solve these equations for the acceleration a.

By the way, when you finish with all this, you will find that you get the same answer that NascentOxygen gave.

Chet
 
Last edited: