Question

for acceleration due to gravity, use $a_{g}=9.81\\ m/s^{2}$; 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 x 10^{23} as 6.02e23 or 6.02x10^23).
problem
answer
1 given: $mu_{s}=0.8$ wanted: angle ($\theta$) in $^{circ}$? angle?
2 given: $f_{p}=11000\\ n$, $a = 0\\ m/s^{2}$, $mu_{s}=0.8$ wanted: mass ($m$) in kg? mass? kg

Explanation:

Step1: Analyze the first problem

We know that the coefficient of static - friction $\mu_s = 0.8$. For an object on an inclined plane on the verge of sliding, $\tan\theta=\mu_s$.

Step2: Calculate the angle

$\theta=\arctan(\mu_s)$. Substituting $\mu_s = 0.8$, we get $\theta=\arctan(0.8)\approx38.7^{\circ}$.

Step3: Analyze the second problem

The net - force formula is $F_{net}=ma$. Given $a = 0\ m/s^{2}$, the frictional force $F_f=\mu_sN$ and the applied force $F = 11000\ N$. Since $a = 0$, the applied force is balanced by the frictional force $F = F_f=\mu_sN$. Also, $N = mg$. So $F=\mu_smg$.

Step4: Solve for mass

We can re - arrange the formula $F=\mu_smg$ to $m=\frac{F}{\mu_sg}$. Given $g = 9.81\ m/s^{2}$ and $\mu_s = 0.8$ and $F = 11000\ N$, then $m=\frac{11000}{0.8\times9.81}\approx1406\ kg$.

Answer:

1. $38.7^{\circ}$
2. $1406\ kg$