It gets attracted due to electro magnetizing
Seafloor spreading<span> is a process that </span>occurs<span> at mid-ocean ridges, where new oceanic crust is formed through volcanic activity and then gradually moves away from the ridge. </span>Seafloor spreading<span> helps explain continental drift in the theory of plate tectonics.
Hope this helps</span>
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
To solve this problem we will use the concepts related to the uniform circular movement from where we will obtain the speed of the object. From there we will go to the equilibrium equations so that the friction force must be equal to the centripetal force. We will clear the value of the coefficient of friction sought.
The velocity from the uniform circular motion can be described as
Here,
r = Radius
T = Period
Replacing,
From equilibrium to stay in the circle the friction force must be equivalent to the centripetal force, therefore
Here,
Coefficient of friction
N = Normal Force
m = mass
v = Velocity
r = Radius
The value of the Normal force is equal to the Weight, then
Rearranging to find the coefficient of friction
Replacing,
Therefore the minimum coefficient of friction to prevent the cat from sliding off is 0.9399
