Answer to Question 2

(a) h = 0.75 m and d = 1.0 m (centre of mass is in the middle)

tan q = h/d = 0.75/1.0 = 0.75

q = tan^-1 (0.75) = 36.9 ^o

But this is from the horizontal.

Angle from the vertical = 90 - 36.9 = 53.1^o (QED)

 

(b) Moment = mg × d sin q = 10000 × 9.8 × 0.75 sin q = 98000 N × 0.75 sin q

Moment from the wind = 5000 N × 3.0 m = 15 000 N m

Þ 15000 = 98000 N × 0.75 sin q

Þ sin q = 15000 ÷ (98000 × 0.75) = 0.204

Þ q = sin^-1 (0.204) = 11.5^o (to the horizontal)

 

(c)  The lorry would not tip over, as 11.5^o is much less than 37^o.