Question

a force of 1760 n is required to start moving a bundle of wooden planks up a ramp. if the ramps incline is 17° and the mass of the planks is 266 kg, what is the coefficient of static friction between the planks and the ramp?

Explanation:

Step1: Analyze forces

When the planks are about to move, the applied force \(F = 1760\ N\) balances the sum of the gravitational force component \(mg\sin\theta\) and the static - friction force \(f_s=\mu_sN\). The normal force \(N = mg\cos\theta\) (from perpendicular - to - the - ramp force balance \(N - mg\cos\theta=0\)).

Step2: Set up force - balance equation

\(F=mg\sin\theta+\mu_smg\cos\theta\). We can solve for \(\mu_s\):

$$
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$$

Given \(m = 266\ kg\), \(g = 9.8\ m/s^2\), \(\theta = 17^{\circ}\), \(F = 1760\ N\)
First, calculate \(mg\sin\theta\): \(mg\sin\theta=266\times9.8\times\sin(17^{\circ})\approx266\times9.8\times0.292\approx763.7\ N\)
Then, calculate \(mg\cos\theta\): \(mg\cos\theta=266\times9.8\times\cos(17^{\circ})\approx266\times9.8\times0.956\approx2507.7\ N\)
Substitute into the \(\mu_s\) formula:

$$
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Answer:

\(\mu_s\approx0.40\)