Question

for acceleration due to gravity, use (g = 9.81 m/s^{2}); ignore air - resistance unless the problem says otherwise. please try to get within 2% of the correct answer. you can enter your answers in decimal notation, or you can enter them in scientific notation using e or x 10^ for the power of 10 (e.g., enter (6.02\times10^{23}) as 6.02e23 or 6.02x10^23)
problem

1. given: wanted: angle ((\theta)) in (circ)?
2. given: (f_x = 11000 n), (a = 0 m/s^{2}), (mu_x = 0.8). wanted: mass (m) in kg?
3. given: (f_x = 450 n), (f_y = 300 n), (a = 0 m/s^{2}). wanted: coefficient of static friction ((mu_s))?
4. given: (m = 6.0 kg), (a = 0 m/s^{2}), (mu_x = 0.6), (\theta = 13^{circ}). wanted: applied force ((f_a)) in n?
5. given: (f_x = 57.0 n), (f_y = 90 n), (a = 0 m/s^{2}). wanted: coefficient of static friction ((mu_s))?

Explanation:

1. Problem 1: Information is insufficient to solve for the angle as no relevant equations - related data is complete.
2. Problem 2:

Step1: Analyze forces in x - direction

Since \(a = 0\ m/s^{2}\), the net force in the x - direction \(F_{net,x}=F_x-\mu_k N = 0\). But we don't have enough information to solve for mass as we don't know the normal force relationship.

1. Problem 3:

Step1: Analyze forces in x - direction

Since \(a = 0\ m/s^{2}\), the net force in the x - direction \(F_{net,x}=F_x - F_y\mu_s=0\). We can re - arrange the equation to solve for \(\mu_s\).

$$F_x - F_y\mu_s = 0\Rightarrow\mu_s=\frac{F_x}{F_y}$$

Step2: Substitute given values

Given \(F_x = 450\ N\) and \(F_y = 300\ N\), then \(\mu_s=\frac{450}{300}=1.5\)

Step1: Analyze forces in x - direction

Since \(a = 0\ m/s^{2}\), the net force in the x - direction \(F_{net,x}=F_a\cos\theta-\mu_k(mg + F_a\sin\theta)=0\).

$$F_a\cos\theta=\mu_k(mg + F_a\sin\theta)$$

$$F_a\cos\theta=\mu_kmg+\mu_kF_a\sin\theta$$

$$F_a\cos\theta-\mu_kF_a\sin\theta=\mu_kmg$$

$$F_a(\cos\theta-\mu_k\sin\theta)=\mu_kmg$$

$$F_a=\frac{\mu_kmg}{\cos\theta-\mu_k\sin\theta}$$

Given \(m = 6.0\ kg\), \(\mu_k = 0.6\), \(\theta = 13^{\circ}\), and \(g = 9.81\ m/s^{2}\)
\(\cos(13^{\circ})\approx0.974\), \(\sin(13^{\circ})\approx0.225\)

$$F_a=\frac{0.6\times6.0\times9.81}{0.974 - 0.6\times0.225}$$

$$F_a=\frac{35.316}{0.974 - 0.135}$$

$$F_a=\frac{35.316}{0.839}\approx42.1\ N$$

Answer:

1.5

1. Problem 4: