Three timbers are placed on a surface in a way that their centers make an equila

Question

Three timbers are placed on a surface in a way that their centers make an equilateral triangle. A constant force F= 1.0 N is then applied lo one of them as shown below. What is the range of acceleration that would allow the timbers to remain in contact? Suggested approach: take timbers as identical cylinders and identify all the forces on each of them. Draw' frce-body diagrams and apply Newton's 2nd law. Solve the system of equations to find the inter-cylinder forces. At the moment the timbers lose contact the forces between them would become zero. Find the range of acceleration that keeps the forces nonzero.

Explanation / Answer

All the three timbers move with accelaration "a" due to force F=1N,this is minimum accelaration needed to keep the timbers in contact because at the moment the timbers lose contact the forces between them would become zero.

suppose the mass of each timber is "m=1kg",then F=3m*a=3a

this leads to acceleration "a"=1/3=.33ms^-2,This is require minimum accelaration.

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