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## Homework Statement

One end of a uniform meter stick is placed against a vertical wall . The other end is held by a lightweight cord that makes an angle θ with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.38. What is the maximum value the angle θ can have if the stick is to remain in equilibrium?

## Homework Equations

F = ma

τ = rFsinφ

## The Attempt at a Solution

I tried doing this problem using the sum of the forces and then the sum of the torques, but I got two different angles that are both incorrect. Here's what I tried:

Using Forces:

x-direction: Fn - Tcosθ = 0 --> T = Fn/cosθ

y-direction: Ff + Tsinθ - Fg = 0

Ff + (Fn/cosθ)sinθ - Fg = 0

μ(Fn) + (Fn)tanθ - Fn = 0

μ + tanθ - 1 = 0

θ = arctan(1-μ) = arctan(1-0.38) = 32°

Using torque:

Taking the reference point at the point between the wall and the stick and taking clockwise rotation to be positive...

τ(stick) + τ(rope) + τ(wall) + τ(friction) = 0

τ(stick) + τ(rope) = 0

τ(stick) = (Fg)(0.5m)sin(90) = 0.5mg

τ(rope) = -(T)(1m)sinθ = -Tsinθ

τ(stick) + τ(rope) = 0.5mg - Tsinθ = 0

Above, we found T = Fn/cosθ

0.5mg - (mg/cosθ)sinθ = 0

0.5 - tanθ = 0

θ = arctan(0.5) = 27°

Neither of these angles are correct. I think I went wrong with the normal force due to the wall, Fn. Is this force equal to the weight of the object? I'm not sure if that makes sense... I think I was wrong to cancel out Fn when I tried using the forces.