Center of mass and tension in two ropes?
A 100kg painter paints the side of a building while standing on a 8m long board. The board is supported by 2 ropes, 1 on each side. If the mass of the board is 25kg and the painter stands 2m from the left end, what are the tensions in each rope?
looked everywhere, couldnt find anything. Final is tomorrow and I just have a feeling this one is going to be on it....
- TechnobuffLv 79 years agoFavorite Answer
Tension in each rope, no painter = (25 x 9.8)/2, = 122.5N.
The painter is 2m. from the left, and therefore 6m. from the right. So there will be (6/2) = 3 x the tension in the left rope than in the right, = 4 parts (3 + 1).
(100 x 9.8) = 980N. painter weight.
(980/4) = 245N. tension in the right rope, and (245 x 3) = 735N. in the left.
Total left = (735 + 122.5) = 857.5N.
Total right = (245 + 122.5) = 367.5N.Source(s): That checks fine.
- 4 years ago
utilising a million/2 the area between the helps you have 2 aspects of a exact perspective triangle. Use pythag to calculate the lenth of the rope decrease than rigidity. stress would be proportional to the the ratio of the aspects. 3480 * sqrt(34.2^2 + (331/2)^2) / 34.2 = 17196.sixteen N
- 6 years ago
sophisticated thing. browse using a search engine. just that might help!