Finding the Normal Force on the Rod of a Paper Roll

1. The problem statement, all variables and given/known data

A large 16.0 kg roll of paper with radius R=18.0 cm rests against the wall and is held in place by a bracket attached to a rod through the center of the roll. The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is 0.260 kg m^2. The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of 30.0° with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is μ= 0.25. A constant vertical force F= 60.0 N is applied to the paper, and the paper unrolls.

(a) What is the magnitude of the force that the rod exerts on the paper as it unrolls?

(b) What is the magnitude of the angular acceleration of the roll?

Here is a picture of that someone else found for the same question, http://ift.tt/PeE8zI

2. Relevant equations

∑Fy = 0

∑Fx = 0

F(reaction) – Tension = 0, so F(rod) = T

F(friction) = μfx(normal)

3. The attempt at a solution

I will try part (a), but this attempt gives me the wrong answer.

∑Fy = T*cos30° – (mg + 60N) = 0

T = (mg + 60N)/cos30° = (16*9.8 + 60N)/cos°30 = 250.3391N

T = F(rod) = 250.3391N.

But the answer is 293N.

I probably need to use the friction, but I don’t understand why.

http://ift.tt/1onHtw8

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