Minimum length x for no slipping

[Ithilrandir]

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

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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.

 

[haruspex]

what is the minimum value of D for no slipping of the hanger?

Taking pivot at the CM,
50W(X - 15) = 2(15)F + 22N

I don't see D defined anywhere.

I think you mean taking a pivot below the CM of the hanger, at the height of the lower bracket.

Where exactly do you have N and F acting on the upper bracket, and which ways do they point?

 

[Ithilrandir]

I don't see D defined anywhere.

I think you mean taking a pivot below the CM of the hanger, at the height of the lower bracket.

Where exactly do you have N and F acting on the upper bracket, and which ways do they point?

The D was a typo, I had meant X. I have F acting on the inner side of the support acting upwards. The upper N is on the left side acting left, the lower N is on the right side acting right.

 

[haruspex]

The D was a typo, I had meant X. I have F acting on the inner side of the support acting upwards. The upper N is on the left side acting left, the lower N is on the right side acting right.

Ok, so you are taking moments about a point above the mass centre, at the height of the upper support. So the friction and normal force in your equation are on the lower support, at the RHS of the column, acting up and to the right respectively.
For each of those, does it act clockwise or anticlockwise about the axis?

 

[Ithilrandir]

Ok, so you are taking moments about a point above the mass centre, at the height of the upper support. So the friction and normal force in your equation are on the lower support, at the RHS of the column, acting up and to the right respectively.
For each of those, does it act clockwise or anticlockwise about the axis?

The Friction acts clockwise, the normal force acts anticlockwise.

So 15F + 50W = 22N.

 

[haruspex]

The Friction acts clockwise, the normal force acts anticlockwise.

So 15F + 50W = 22N.

What happened to X?

 

[Ithilrandir]

What happened to X?

My bad. 15F + 50(X-15) = 22N. Is this right?

 

[haruspex]

My bad. 15F + 50(X-15) = 22N. Is this right?

Looks good.

 

[Ithilrandir]

My bad. 15F + 50(X-15) = 22N. Is this right?

This doesn't seem right in calculations as I get 10.716 for X.

 

[haruspex]

This doesn't seem right in calculations as I get 10.716 for X.

Yes, I missed another error.
Both frictional forces need to be in that torque balance equation.

 

[Ithilrandir]

Yes, I missed another error.
Both frictional forces need to be in that torque balance equation. Or consider them as a couple, which means the distance is not 15cm.

The two frictional forces should be in the same direction, so isn't the net effect just 2F x 15?

 

[haruspex]

The two frictional forces should be in the same direction, so isn't the net effect just 2F x 15?

They act on opposite sides of the shaft.
(Strike my remark about a couple - must be past my bedtime.)

 

[Ithilrandir]

They act on opposite sides of the shaft.
(Strike my remark about a couple - must be past my bedtime.)

Ah I see. I should've placed the friction where I had placed the normal force. I will test it tmr after work.

 

[Ithilrandir]

I manged to get the answer after pouring fixing some mistakes in my algebra with the new equation.