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. the ramp makes an angle of \\( \theta = 37.5 ^ { \circ } \\) with the horizontal. the package has a mass of 9.75 kg. the coefficient of static friction between the ramp and the package is 0.405. calculate the minimum force \\( f _ { \min } \\) that this arm must apply to hold a package on the ramp. \\( f _ { \min } = \\) n

Explanation:

Step1: Resolve forces

The gravitational force \(mg\) can be resolved into components. The component along the ramp is \(mg\sin\theta\) and the component perpendicular to the ramp is \(mg\cos\theta\). Let the normal force be \(N\) and the horizontal force be \(F\).
Resolving forces perpendicular to the ramp: \(N = mg\cos\theta+F\sin\theta\)
Resolving forces along the ramp: \(mg\sin\theta=F\cos\theta + f_s\), where \(f_s=\mu_sN\) (static - friction force, \(\mu_s\) is the coefficient of static friction)

Step2: Substitute \(N\) into the along - ramp force equation

Substitute \(N = mg\cos\theta+F\sin\theta\) into \(mg\sin\theta=F\cos\theta+\mu_sN\)

$$mg\sin\theta=F\cos\theta+\mu_s(mg\cos\theta + F\sin\theta)$$

$$mg\sin\theta=F\cos\theta+\mu_smg\cos\theta+\mu_sF\sin\theta$$

$$mg\sin\theta-\mu_smg\cos\theta=F(\cos\theta+\mu_s\sin\theta)$$

Step3: Solve for \(F\)

$$F=\frac{mg(\sin\theta-\mu_s\cos\theta)}{\cos\theta+\mu_s\sin\theta}$$

Given \(m = 9.75\space kg\), \(g = 9.8\space m/s^{2}\), \(\theta = 37.5^{\circ}\), \(\mu_s=0.405\)
\(\sin(37.5^{\circ})\approx0.609\), \(\cos(37.5^{\circ})\approx0.793\)

$$F=\frac{9.75\times9.8\times(0.609 - 0.405\times0.793)}{0.793+0.405\times0.609}$$

First, calculate the numerator:

$$9.75\times9.8\times(0.609 - 0.405\times0.793)=9.75\times9.8\times(0.609 - 0.321)=9.75\times9.8\times0.288$$

$$9.75\times9.8 = 95.55$$

, \(95.55\times0.288 = 27.52\)
Then, calculate the denominator:

$$0.793+0.405\times0.609=0.793 + 0.247=1.04$$

$$F=\frac{27.52}{1.04}\approx26.5$$

Answer:

\(26.5\)