2007-11-26
Today's nonsense
- Seniors have "passes" to come in late on Wednesday.
Unofficially starting a review for the test. Test is Thursday/Friday.
Homework #14 - Rock band, Dead Weight, the band starts off with the sound of an automobile accident, and the lead singer slides out from the back with some extreme guitar moves that go BAM BAM BAM. So, lead singer, Sharika, goes to the floor and slides on her knees, with an initial speed of 3 m/s (V_o) and after sliding for d = 2 m, and she rests in between flashpots. What's the coefficient of kinetic friction between Sharika and the stage?
Free body diagram. Maybe. Absolutely.
F_n - mg = 0
f_k = mu sub k * F_n =mu sub k * mg
Sum of F_y = F_n - mg = 0
Sum of F_x = -f_k = ma
Your f_k = -ma, so
-ma = mu_k * mg
-a = mu_k * g
mu_k = -a / g
Now, the acceleration, they had an initial velocity and final velocity, so we can do our final velocity squared = initial velocity squared + 2ad
Final velocity = 0, so:
0 = 3^2 + 2a(2)
So that's -9/4 = a
So: mu_k = -(-9/4)/9.81 = 0.229
#15) We have a 5 kg block that is held against the wall with a horizontal force of 100 Newtons.
Sum of F_x = F_n - F = 0 (because it's holding it against the wall and the acceleration is therefore zero)
Sum of F_y = f_s - mg = 0
Part A) Find the friction force, and that just equals mg so f_s = mg = 49.05 N
Part B) What's the minimum horizontal force to prevent the block from falling if the coefficient of friction is 0.4?
mu_s = 0.4
f_s = mu_s * F_n
F_n = f_s / mu_s = 49.05 / 0.4 = 122.625 N
#16) We have an angle.. We did this problem as far as angle and the mu_s as relationship between mu_s and the angle. You can probably find this in your notes but we'll rederive it.
On a snowy day with the temperature near the freezing point, the coefficient of static friction between the car's tires and the road is 0.8. And it says what's the maximum incline that the four-wheel drive vehicle can climb with zero acceleration?
Sum of F_x = f_s - mgsin(theta) = 0
Sum of F_y = F_n - mgcos(theta) = 0
We are trying to solve for theta.
f_s = mu_s * F_n
Say that f_s = mu_s * F_n = mu_s * mgcos(theta) = mgsin(theta)
mu_s = mgsin(theta) / mgcos(theta) = sin(theta)/cos(theta)
mu_s = tan(theta)
theta = arctan(mu_s)
theta = 4.574 degrees
#18) A 3 kg resting on a horizontal shelf is connected to a 2 kg block by a light string. What is the mimum coefficient of static friction (mu_s) so that the two objects remain at rest?
mu_s = ? so that a = 0
f_s = T = 2g = mu_s * F_n = mu_s * 3g so that mu_s = 2/3
Suppose the actual mu_s is less than the one they actually gave us, so that means that we'd actually be moving and we'd have a coefficient of kinetic friction. Find the time for the 2kg mass to fall two meters.
mu_k = 0.3
So the whole issue with falling 2 meters, we don't care about that yet, we have to find the acceleration, and once you find the acceleration you can do the whole velocity squared = velocity squared initial + 2ad and then we'd do our sort of s = 1/2 at^2 + initial velocity * t plus initial position or something.
Sum of F_x = T - f_k = ma = 3a
Sum of F_y = F_n - 3g = 0 because in the y-direction it is not moving
Then on the other side:
Sum of F_y = T - 2g = 2(-a)
a_1 = -a_2 = a
So solve for a. Basically you should realize that you have to solve for T.
T = T
3a + f_k = 2g - 2a
3a + mu_k * F_n = 3a + mu_k * 3g = 2g - 2a
5a + mu_k * 3g = 2g
5a = 2g - 0.3(3g)
a = 1.1g/5 = 2.1582 m/sec^2
How long is it going to take that block to fall 2 meters?
S = (1/2)aT^2 + V_o * T + S_o
2 m = (1/2)(2.1582)T^2 + 0T + 0 meters
4 = 2.1582 * T^2
4/2.1582 = T^2
T = sqrt(4/2.1582) = 1.361 seconds
19)
Sum of F_x = -f_k = ma
Sum of F_y = F_n - mg = 0
f_k = ma
mu_k = -ma / mg ???
f_k = mu_k * F_n
-ma = mu_k * mg
mu_k = -ma/mg
mu_k = -a/g
That answer is not good enough. You know your velocity, and you need to find your a in terms of v and d. And in this scenario V_f = 0
0^2 = V_o squared + 2ad
0 = V^2 + 2ad