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
Because it gives it more space to be able to move farther and faster which can change the speed in which you want it to go
A vector quantity is defined by magnitude and direction. For example, we might say that a car has an average speed of 25 miles per hour. Its average velocity might be 25 miles per hour due east.
Answer:
its "up, then down, then up"
Explanation:
Answer:
3.5
Explanation:
The angular magnification, M = 1 + 25cm/f where f = focal length = 10.0 cm
M = 1 + 25cm/f
= 1 + 25cm/10.0 cm
= 1 + 2.5
= 3.5
