mclame22
Nov13-10, 05:42 PM
1. The problem statement, all variables and given/known data
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?
2. Relevant equations
F = ma
τ = rFsinφ
3. 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.
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?
2. Relevant equations
F = ma
τ = rFsinφ
3. 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.