I believe another name for lower point is Ice point
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.
1) Equilibrium of forces:

where

is the weight of the person

is the weight of the scaffold
Re-arranging, we can write the equation as

(1)
2) Equilibrium of torques:

where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using

and replacing T1 with (1), we find

from which we find

And then, substituting T2 into (1), we find
Answer:
largest lead = 3 m
Explanation:
Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.
So, first we need to calculate the velocities of both the anchorman
given data:
Distance = d = 100 m
Time arrival for A = 9.8 s
Time arrival for B = 10.1 s
Velocity of anchorman A = D / Time arrival for A
=100/ 9.8 = 10.2 m/s
Velocity of anchorman B = D / Time arrival for B
=100/10.1 = 9.9 m/s
As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use
d = vt
= 9.9 x 9.8 = 97 m
So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.
largest lead = 100 - 97 = 3 m
So if his lead no more than 3 m anchorman A win the race.
Answer:
the pressure at the depth is 1.08 ×
Pa
Explanation:
The pressure at the depth is given by,
P = h
g
Where, P = pressure at the depth
h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter
= density of water = 1000 
g = acceleration due to gravity ≈ 9.8 
P = 11033.15 × 9.8 × 1000
P = 1.08 ×
Pa
Thus, the pressure at the depth is 1.08 ×
Pa