physics----equilibrium

2009-01-23 6:25 pm
1. A 36kg round table is supported by three legs placed equal distances apart on the edge. What minimum
mass, placed on the table's edge, will cause the table to overturn??




2. A 160kg horizontal beam is supported at each end. A 300 kg piano rests a quarter of the way from one end. What is the vertical force on each of the supports?
更新1:

according to your answer for question 1, the question doesn't mention the diameter of the table, leg distance from the edge and the position of the Mass relative to the legs. I just copied the question right away.....

回答 (2)

2009-01-24 7:21 pm
✔ 最佳答案
1. You have to draw a geometrical diagram as follows:
Join the three legs of the table together to form a large triangle. From the centre of the table (where the centre of gravity is located), draw lines joining the table centre with the three legs.
There will then be three smaller isosceles triangles forming. The apex of each triange is 360/3 degrees = 120 degrees. Take any one triangle, drop a perpendicular line from the apex (i.e. the table centre) to the base of the triangle and produce it to meet the rim of the table at a point, say point A. This perpenicular line would divide the triangle into two identical ones, each with apex angle 120/2 deg. = 60 degrees.
The required mass (M, say)is placed at A to topple the table. The pivot is the line joining the nearest two legs from A (i.e. the base line of the triangle)
Distance of the table centre from the pivot
= R.cos(60)
where R is the radius of the table
Moment given by the table mass = 36 x R.cos(60)
Distance of A from the pivot = R-R.cos(60)
Hence, moment given by the mass at A = M.[R-R.cos(60)]
For equilibrium,
36 x R.cos(60) = M.[R-R.cos(60)]
M = 36.cos(60)/(1-cos(60)) kg = 36 kg
------------------------------------------
2. Let the end nearest to the piano be A and the other end be B. The reactions at A and B are R and R' respectively
hence, R + R' = (160 + 300)g N = 460g N
where g is the acceleration due to gravity
i.e. R = 460g - R'
Take moment about A
R'.L = 160g(L/2) + 300g(L/4)
where L is the length of the beam
R = (160/2 + 300/4)g N = 155g N = 1550 N (take g = 10 m/s2)
thus, R' = (460-155)g N = 305g N = 3050 N
2009-01-24 6:00 am
1. Not sufficient information.
To calculate it, need following information:
a. diameter of the table.
b. leg distance from the edge.
c. position of the Mass relative to the legs.

2.
At the end closer to the piano:
(1/2)160+(3/4)300 = 305kg

At another end:
(1/2)160+(1/4)300 = 155kg

2009-01-26 22:12:41 補充:
002,
How about placing the Mass right above one of the legs?
What assumptions you have made?


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