Since the ladder is standing, we know that the coefficient
of friction is at least something. This [gotta be at least this] friction
coefficient can be calculated. As the man begins to climb the ladder, the
friction can even be less than the free-standing friction coefficient. However,
as the man climbs the ladder, more and more friction is required. Since he
eventually slips, we know that friction is less than what's required at the top
of the ladder.
The only "answer" to this problem is putting lower
and upper bounds on the coefficient. For the lower one, find how much friction
the ladder needs to stand by itself. For the most that friction could be, find
what friction is when the man reaches the top of the ladder.
Ff = uN1
Fx = 0 = Ff + N2
Fy = 0 = N1 – 400 – 864
N1 = 1264 N
Torque balance
T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)
N2 = 439 N
Ff = 439= u N1
U = 440 / 1264 = 0.3481
Answer:
The ball will hit the ground in 2.73 s
Explanation:
Hi there!
f(t) is the height of the ball at time "t". We need to find the value of "t" for which f(t) = 0.
f(t) = 0 Then:
-16 · t² + 40 · t + 10 = 0
Solving the quadratic equation using the quadratic formula (a = -16, b = 40, c = 10):
t = -0.23 s and t = 2.72 s
Since time can´t be negative, be discard that value.
The ball will hit the ground in 2.73 s.
Have a nice day!