Question

a crate is pushed across a level floor by a force of $3.00\times10^{2} n$ exerted at an angle of $20.0^{circ}$ below the horizontal. the coefficient of kinetic friction between the crate and floor is $0.250$. if the crates velocity is constant, what is the magnitude of the normal force exerted on the crate by the floor? what is the mass of the crate?

Explanation:

Step1: Analyze the vertical forces

Since the crate's velocity is constant, the net force in the vertical direction is zero. The vertical forces are the normal force \(N\) (upward), the vertical component of the applied force \(F_y = F\sin\theta\) (downward), and the weight \(mg\) (downward). So, \(N=F\sin\theta + mg\).

Step2: Analyze the horizontal forces

In the horizontal direction, the net force is also zero (constant velocity). The horizontal component of the applied force \(F_x=F\cos\theta\) is balanced by the kinetic - friction force \(f_k\). The kinetic - friction force \(f_k=\mu_kN\). So, \(F\cos\theta=\mu_kN\).

Step3: Solve for the normal force \(N\)

From \(F\cos\theta=\mu_kN\), we can express \(N=\frac{F\cos\theta}{\mu_k}\). Given \(F = 3.00\times10^{2}\text{ N}\), \(\theta = 20.0^{\circ}\), and \(\mu_k=0.250\).

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$$N=\frac{281.91}{0.250}=1127.64\text{ N}$$

Step4: Solve for the mass \(m\)

From \(N=F\sin\theta+mg\), we can solve for \(m\). First, find \(F\sin\theta\):

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Then, \(mg=N - F\sin\theta\). Substitute \(N = 1127.64\text{ N}\) and \(F\sin\theta=102.6\text{ N}\)

$$mg=1127.64 - 102.6=1025.04\text{ N}$$

Since \(g = 9.8\text{ m/s}^2\), \(m=\frac{1025.04}{9.8}=104.6\text{ kg}\)

Answer:

The magnitude of the normal force \(N = 1128\text{ N}\) and the mass of the crate \(m = 105\text{ kg}\)