Question

010 (part 1 of 3) 10.0 points
a block is at rest on the incline shown in the figure. the coefficients of static and ki - netic friction are $mu_s = 0.67$ and $mu_k = 0.57$, respectively.
the acceleration of gravity is $9.8 m/s^2$.
what is the frictional force acting on the 17 kg mass?
answer in units of n.
011 (part 2 of 3) 10.0 points
what is the largest angle which the incline can have so that the mass does not slide down the incline?
answer in units of $^{circ}$.
012 (part 3 of 3) 10.0 points
what is the acceleration of the block down the incline if the angle of the incline is $38^{circ}$?
answer in units of $m/s^2$.

Explanation:

Step1: Analyze forces for part 1

Since the block is at rest, the frictional force $f$ balances the component of the gravitational - force along the incline. The component of the gravitational force along the incline is $F_{g\parallel}=mg\sin\theta$, where $m = 17\ kg$, $g=9.8\ m/s^{2}$, and $\theta = 31^{\circ}$.
$f = mg\sin\theta=17\times9.8\times\sin31^{\circ}$

Step2: Calculate the value for part 1

$f=17\times9.8\times0.515=17\times5.047 = 85.799\ N\approx85.8\ N$

Step3: Analyze forces for part 2

At the maximum - angle $\theta_{max}$ before the block starts to slide, the static - frictional force reaches its maximum value $f_s=\mu_sN$. The normal force $N = mg\cos\theta_{max}$, and the component of the gravitational force along the incline is $mg\sin\theta_{max}$. When the block is on the verge of sliding, $mg\sin\theta_{max}=\mu_smg\cos\theta_{max}$. Then $\tan\theta_{max}=\mu_s$.
$\theta_{max}=\arctan(\mu_s)$

Step4: Calculate the value for part 2

$\theta_{max}=\arctan(0.67)\approx33.8^{\circ}$

Step5: Analyze forces for part 3

The net force along the incline is $F_{net}=mg\sin\theta-\mu_kN$. The normal force $N = mg\cos\theta$. According to Newton's second law $F_{net}=ma$, so $mg\sin\theta-\mu_kmg\cos\theta=ma$. Then $a = g(\sin\theta-\mu_k\cos\theta)$.
$a = 9.8(\sin38^{\circ}-0.57\cos38^{\circ})$

Step6: Calculate the value for part 3

$\sin38^{\circ}\approx0.616$, $\cos38^{\circ}\approx0.788$.
$a = 9.8(0.616 - 0.57\times0.788)=9.8(0.616-0.44916)=9.8\times0.16684\approx1.63\ m/s^{2}$

Answer:

010: 85.8 N
011: 33.8°
012: 1.63 m/s²