A stick held against a wall by a rope?

[mclame22]

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.

 

[mark726]

Any help would be appreciated!Your approach to the problem is correct, but there are a few errors in your calculations.

First, in the x-direction, the tension force should be Tsinθ, not Tcosθ. This is because the tension force is acting at an angle θ from the horizontal, not the vertical.

Second, in the y-direction, the force of friction should be μFn, not μFg. This is because the force of friction is between the wall and the stick, not between the stick and the ground.

Third, in your torque calculation, you have used the wrong distance for the stick. The distance from the reference point to the center of mass of the stick is not 0.5m, it is 0.5m*cosθ. This is because the stick is not perpendicular to the wall, it is at an angle θ.

With these corrections, your calculations should give the correct answer. Keep in mind that the normal force, Fn, is not equal to the weight of the object in this case. It is equal to the component of the weight that is perpendicular to the wall, which is mg*cosθ.