36;in other words,(7)tan(θ)=1 / 1.36 =0.735⇒(8)θ ~= 36 degrees .-A uniform board is leaning against a smooth vertical wall. The board is at an angle above the horizontal ground.......
= 2 * H / Fn
Finally, using (4):
= 2 * μs = 2 * 0.680 = 1.36; in other words,
(7) tan(θ) = 1 / 1.36 = 0.735 ⇒
(8) θ ~= 36 degrees <<===Answer
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A uniform board is leaning against a smooth vertical wall. The board is at an angle above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is 0.680. Find the smallest value for the angle , such that the lower end of the board does not slide along the ground.
The weight of the board is pulling down at the center of the board. If the board is L meters long, the center is at L/2.
The vertical wall has no friction, so the ground is supporting the entire weight of the board.
The friction between the ground and the lower end of the board is keeping the lower end from moving farther away from bottom of the wall.
Let the pivot point the point where the top of the ladder meets the wall.
The weight times perpendicular distance pivot point = torque
The perpendicular distance from pivot point = L/2 * cos θ
Torque = m* g * L/2 * cos θ
The friction force times perpendicular distance pivot point = torque in opposite direction
The perpendicular distance from pivot point = L * sin θ
The ground is exerting an upward force equal to the weight = m* g
Friction force is horizontal = μ * upward force = μ * * m * g
Opposite Torque = μ * m * g * L * sin θ
μ * m * g * L * sin θ = m* g * L/2 * cos θ
divide both sides by m * g * L
μ * sin θ = ½ * cos θ
sin θ ÷ cos θ = ½ ÷ μ
tan θ = ½ ÷ 0.680
Find θ
It's a long time since I was at school and this question gave me a massive headache :0
