A house is built on the top of a hill with a nearby slope at angle 45° slope (Figure 6-55). An engineering study indicates that the slope angle should be reduced because the top layers of soil along the slope might slip past the lower layers. If the static coefficient of friction between two such layers is 0.42, what is the least angle ϕ through which the present slope should be reduced to prevent slippage?
Slippage happens when the tangent of the slope angle exceeds the static coefficient of friction. Use (or derive) that fact to figure out the answer
A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.00 g coins stacked over the 12.0 cm mark, the stick is found to balance at the 43.0 cm mark. What is the mass of the meter stick?
Set the total moment about the knife edge equal to zero.
(meter mass)*(center distance from knife edge) = (coin mass)* (distance from knife edge)
(43 - 12)* m = 10.00 g * 7.0
Solve for m
To find the least angle ϕ through which the present slope should be reduced to prevent slippage, we need to understand the concept of static friction.
The static friction force between two surfaces is given by the equation:
F_f = μ_s * N
where F_f is the friction force, μ_s is the static coefficient of friction, and N is the normal force between the two surfaces.
In this case, the normal force N can be divided into two components: one perpendicular to the slope (N_perp) and one parallel to the slope (N_parallel).
To prevent slippage, the friction force acting parallel to the slope needs to be equal to or greater than the force that would cause the upper layers of soil to start sliding down the slope.
The force that would cause slippage can be calculated by multiplying the weight of the upper layers of soil by the sine of the slope angle:
F_slippage = W * sin(45°)
where W is the weight of the upper layers of soil.
The friction force acting parallel to the slope can be calculated by multiplying the normal force parallel to the slope (N_parallel) by the static coefficient of friction:
F_friction = μ_s * N_parallel
Since N_parallel is equal to N * sin(ϕ), where ϕ is the angle through which the slope should be reduced, we can substitute N_parallel into the equation:
F_friction = μ_s * N * sin(ϕ)
To prevent slippage, the friction force must be equal to or greater than the force that would cause slippage:
μ_s * N * sin(ϕ) ≥ W * sin(45°)
We can simplify this equation by dividing both sides by sin(ϕ):
μ_s * N ≥ (W * sin(45°)) / sin(ϕ)
Now we can solve for the angle ϕ:
ϕ ≥ arcsin((W * sin(45°)) / (μ_s * N))
Therefore, the least angle ϕ through which the present slope should be reduced to prevent slippage is equal to or greater than arcsin((W * sin(45°)) / (μ_s * N)).