Question

1. a force of 200 n is just sufficient to start a 50 kg steel trunk moving across a wooden floor. find the coefficient of static friction.

n = 500 n
f_{q}=200 n
mg=(50)(10)=500
200 = \mu\cdot500
\frac{200}{500}=\frac{\mu\cdot500}{500}

Explanation:

Step1: Calculate the normal force

The normal force \(N\) on a horizontal surface is equal to the weight of the object. Using the formula \(N = mg\), where \(m = 50\space kg\) and \(g= 10\space m/s^{2}\).
\(N=50\times10 = 500\space N\)

Step2: Use the formula for static - friction force

The maximum static - friction force \(F_{s}=\mu_{s}N\). Given that the force required to start the motion \(F = 200\space N\) (which is equal to the maximum static - friction force \(F_{s}\)).
We have \(200=\mu_{s}\times500\)

Step3: Solve for the coefficient of static friction \(\mu_{s}\)

Rearrange the equation \(200=\mu_{s}\times500\) to get \(\mu_{s}=\frac{200}{500}\)
\(\mu_{s}=0.4\)

Answer:

The coefficient of static friction \(\mu_{s}\) is \(0.4\)