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- Homework Statement
- Two identical rough cylinders of radius r and weight W rest, not touching each other but a negligible distance apart, on a horizontal floor. A thin flat rough plank of width 2a, where a < r, and weight kW rests symmetrically and horizontally on the cylinders, with its length parallel to the axes of the cylinders and its faces horizontal. A vertical cross-section is shown in the diagram below.

The coefficient of friction at all four contacts is 1/2. The system is in equilibrium. 2

(i) Let F be the frictional force between one cylinder and the floor, and let R be the normal reaction between the plank and one cylinder. Draw a free body diagram and show that

Fsin(theta) = R(1+cos(theta))

- Relevant Equations
- Net torque = 0, Net force in horizontal direction = 0.

I have put a picture of the problem above. The picture below shows my free body diagram. Fg is the component of the gravitational force in the direction fo the tangent at the point of contact and Ff is the frictional force between the plank and cylinder. I put Ff in the opposite direction as Fg since the plank naturally wants to slide downwards.

The answer has this free body diagram:

I understand the equations (you got to use the fact that the net torque and net force in the horizontal direction is zero), but I do not know why in the answer diagram F1 (which is the frictional force between the plank and cylinder) is what causes the anti clockwise rotation instead of Fg (which in my diagram represent the component of the gravitational force on the plank which is in the direction of the tangent at the point of contact). Also, I don't understand how the direction of R was chosen in the answer diagram. If somebody can help, pose do so. Thanks.