Firstly for part a, we find the AC moment of the 450N 40cm away from the pivot by doing 40cos30 x 450 (to get 15588.5Ncm)

Next we find the way the vertrical distance of F, so we do 25sin15 (to get 6.47cm)

Finally we 450 x 40cos30 / 25sin15 to get 2409N. This is part i. For part ii, replace 40cos30 with 40cos70 to get 951N

For part b we need to recognise and state:

1. Larger force with smaller angle

2. AC moment large as perpendicular distance to pivot is large

3. The best position would be to bend knees and keep the load close to your body.

(essentially we are keeping the distance as short as possible).

	Created by Doc Boff about 8 years ago

Difficult exam question merging COM and moments