Verified answer

The mans weight under normal; conditions (at rest ) is m*g = 100*9.81 = 981 N

If the scale says 1960N that means there is a net force on the scale of 1960 - 981 = 979 N

Newton's Law, F = ma

a = F/m = 979 / 100 = 9.79 m/s^2

Forces acting on our guy - his weight (-m*g) and the normal force (N = 1960 Newton).

N - m*g = m*a

a = N/m - g = 1960 Newton / 100 Kg - 9.8 m/s^2 = 1.96 m/s^2 - 9.8 m/s^2 = 9.8 m/s^2

The gravitational rigidity continually is directed (very virtually) in the direction of the middle of the Earth. with the aid of fact the truck tips, the line of gravitational rigidity continues to be directed down. on the venture of the truck, the rigidity of gravity strikes to the left with the aid of fact the truck tips. it somewhat is the comparable venture as though the truck have been turning a magnificent hand corner. If the line of rigidity is going exterior of the wheelbase, it is going to tip over. to that end once you drop the line of rigidity from the middle of mass to the exterior edge of the left tire, it types a triangle. the attitude on the CG element is comparable to theta. So the legs of the triangle are the peak of CG (2.2m) and one million/2 the width of the truck (one million.2m). you comprehend that tan(theta)=opposite/hypotenuse=one million.2m/2.2m... to that end theta=atan(one million.2/2.2)

mg = 100 x 9.8 = 980 N

1960/100 = (g + a) = 19.6

19.6 - g = a

19.6 - 9.8 = 9.8 m/s^2