Solving Spring CD Horizontal Position Equilibrium Problem

[musicmar]

Homework Statement

This is actually an engineering problem, but no-one ever seems to answer my questions when I post them there, and this is physics related.

Spring CD remains in the horizontal position at all times due to the roller at D. If the spring is unstretched when θ= 0° and the stiffness is k = 1.5 kN/m, determine the smallest angle θ for equilibrium and the horizontal and vertical components of reaction at pin A.
I've attached the picture. Sorry for the terrible image quality, but I think it's readable.

The Attempt at a Solution

Use equilibrium equations: unknowns: A_x, A_y, θ

F_DC=kx= (1.5)(0.6cosθ) = 0.9cosθ

∑F_y=0= -300 + A_y
A_y= 300 N

∑F_x=0= A_x-F_DC
A_x=0.9cosθ

∑M_A=0= (-300)(0.6) + F_DC(0.6sinθ)
0= -18 + 0.54sinθcosθ
33.33=sinθcosθ

So, the problem is that sinθcosθ can't equal 33.33. Also, for the level of this class, I don't think I should have a trig product like this. Everything else we've done can be solved with basic algebra.

 

[ehild]

Look after the definition of torque in your textbook. ehild

 

[musicmar]

A moment is defined as the force x perpendicular distance to axis to the line of action of the force. F_CD is defined by the angle because no horizontal dimensions are given. Then the perpendicular distance and thus the moment about A due to this force is dependent on the sine of the angle.

 

[ehild]

What is the direction of Fdc? How long is the perpendicular distance of line of action from the axis of rotation at the pin?
The same for the 300 N force. What is the distance of its line of action from the pin?

ehild

 

[musicmar]

It depends on where the axes are defined. If the origin is at point A, oriented with x horizontally, the perpendicular distance from A to DC is (0.6)sinθ. If the axes are rotated so that the x-axis runs along AC, then the perpendicular distance is 0.6 m, and only the x-component causes a moment. This would simplify this part, but it appears that it would complicate other parts of the problem.

 

[ehild]

The perpendicular distance from A to DC is not (0.6)sinθ. By the way, how is sinθ defined?

The rotation happens in the plane of the picture, the axis of rotation is normal to that plane. There is no sense to find torque along an axis in-plane. First try to understand what "torque" means. See for example http://www.education.com/reference/article/rotational-equilibrium/?page=2.

ehild

 

[musicmar]

I found another mistake I made. k is given in kN/m, not N/m.

 

[ehild]

yes, k= 1500 N/m. Have yyou found the level arm of F?

This is not a simple problem, and you can not solve it with simple algebra.

ehild

 

[musicmar]

I had to turn it in with what I had. It worked and looked reasonable in the end. Thank you for your help.

 

[ehild]

The distance of the line of action of F from A is 0.6 sin(θ) + 0.45 cos (θ). This yields the following equation for the torque

-300 (0.6 sin(θ) + 0.45 cos (θ))+1500*0.36 sin(θ)cos(θ)=0

The equation can be solved numerically, giving θ=27.6 degrees.

ehild