Question

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 33.5° with the horizontal. The package has a mass of 9.25 kg. The coefficient of static friction between the ramp and the package is 0.455.

Answer in _____N

Answer 1

To find the minimum force needed by a spring-loaded arm to hold a package on a ramp, calculate the normal force, the force of friction, and the component of gravity acting down the ramp. The sum of the force due to gravity and the force of friction equals the minimum force needed to hold the package in place, which is approximately 84.47 N.

Step-by-step explanation:

To calculate the minimum force that must be applied by a spring-loaded arm to hold a package on a ramp, we need to analyze the forces acting on the package. We are given a package with mass 9.25 kg, a ramp inclined at 33.5°, and a static coefficient of friction of 0.455.

The force due to gravity acting down the ramp can be calculated using the component of the weight of the package that is parallel to the ramp: Fg,parallel = mgsin(θ), where m is the mass, g is the acceleration due to gravity (9.8 m/s2), and θ is the angle of the ramp. The force of friction, which resists this motion, is Ffriction = μsN, where μs is the static coefficient of friction, and N is the normal force exerted by the ramp on the package. The normal force can be found by N = mgcos(θ).

Now we can find the horizontal force that needs to be applied by the spring-loaded arm to hold the package in place. The minimum force required is the sum of the force due to gravity acting down the ramp and the force of friction. So, the calculation will be:

1. Calculate the normal force: N = mgcos(θ).
2. Calculate the force of friction: Ffriction = μsN.
3. Calculate the component of gravitational force down the ramp: Fg,parallel = mgsin(θ).
4. Sum the component of gravitational force and the force of friction to find the stopping force: Ftotal = Fg,parallel + Ffriction.

Plugging in the given values:

• N = (9.25 kg)(9.8 m/s2)cos(33.5°)
• Ffriction = 0.455N
• Fg,parallel = (9.25 kg)(9.8 m/s2)sin(33.5°)

Performing these calculations:

• N ≈ 75.70 N
• Ffriction ≈ 34.44 N
• Fg,parallel ≈ 50.03 N
• Ftotal = 50.03 N + 34.44 N ≈ 84.47 N

Therefore, the minimum force that must be applied by the arm is approximately 84.47 N.