Question

At a mail sorting facility, packages slide down a ramp but are stopped part way down...

At a mail sorting facility, packages slide down a ramp but are stopped part way down the ramp so that they can be scanned. While the packages are scanned, they are held in place by a horizontal force from a spring‑loaded arm. Calculate the minimum force that must be applied by this arm to hold a package on the ramp. The ramp is at an angle of ?=31.5∘θ=31.5∘ with the horizontal. The package has a mass of 9.25 kg.9.25 kg. The coefficient of static friction between the ramp and the package is 0.405.

Solution:

The force pulling the package down the ramp is the component of the weight in the direction of the ramp: Fw = m*g*sinθ

The component of the spring force in the direction of the ramp is Fs = Fa*cosθ.

Before the package starts to move, this force will be aided by the frictional force, which is Fr = µ*Fn, where Fn is the net normal force of the package against the ramp:

Fn = m*g*cosθ + Fa*sinθ

then, Fr = µ*(m*g*cosθ + Fa*sinθ)

For the package to slide, the weight must at least equal the applied force component and overcome the friction force

m*g*sinθ = Fa*cosθ + µ*(m*g*cosθ + Fa*sinθ)

Fa*(cosθ + µ*sinθ) = m*g*(sinθ - µ*cosθ)

Fa = m*g*(sinθ - µ*cosθ)/(cosθ + µ*sinθ)

Using the known values,

Fa = 9.25 * 9.8 * (Sin 31.5 - 0.405*Cos 31.5)/(Cos 31.5 + 0.405*Sin 31.5)

Fa = 90.74 * (0.177)/(1.064)

Fa = 15.10 Newton