- #1
Ithilrandir
- 82
- 3
- Homework Statement
- Adjustable supports that can be slid up and down vertical posts are very useful in many applications. Such a support is shown in the figure, with pertinent dimensions. If the coefficient of static friction between post and support is 0.30, and if a load 50 times the weight of the hanger is to be placed on the hanger at X, what is the minimum value of D for no slipping of the hanger?
- Relevant Equations
- ...
I'm letting the weight of the hanger be W.
Since there is no slipping, the total frictional force will be = total weight.
When the load of 50W is placed at X, there'll be a normal force at the left end of the pole on top to the left, and another normal force at the right end of the pole at the bottom to the right.
Since friction F = 0.30 normal force N,
Taking pivot at the CM,
50W(X - 15) = 2(15)F + 22N
Total weight = 51W, and since there are two frictional forces (top and bottom),
2F = 51W,
50WX - 750W = 765W + 168.3W
X= 33.666 cm.
The answer is 32cm, so there is something wrong with my steps.
Since there is no slipping, the total frictional force will be = total weight.
When the load of 50W is placed at X, there'll be a normal force at the left end of the pole on top to the left, and another normal force at the right end of the pole at the bottom to the right.
Since friction F = 0.30 normal force N,
Taking pivot at the CM,
50W(X - 15) = 2(15)F + 22N
Total weight = 51W, and since there are two frictional forces (top and bottom),
2F = 51W,
50WX - 750W = 765W + 168.3W
X= 33.666 cm.
The answer is 32cm, so there is something wrong with my steps.