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:

Explanation:
As we know that magnitude of two vectors is given as

here we know that
A = magnitude of vector A
B = magnitude of vector B
= angle between two vectors
so here we know that
A = 30 units
B = 40 units
angle = 90 degree
so we have



Answer:
(a) 61.25 N
(b) 6.25 kg
(c) 6.25 Kg
Explanation:
Weight on moon = 10 N
Acceleration due to gravity on moon = 1.6 m/s^2
Acceleration due to gravity on earth = 9.8 m/s^2
Let m be the mass of the package.
(a) Weight on earth = mass x acceleration due to gravity on earth
Weight on earth = 6.25 x 9.8 = 61.25 N
(b) Weight on moon = mass x acceleration due to gravity on moon
10 = m x 1.6
m = 6.25 kg
(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.
Answer: Construction management, Architecture, Civil Drafting.
Explanation: hope this helps :)