Torque and static equilibrium problem check

[Aristotle]

Homework Statement

Homework Equations

Static equilibrium: Net force & net torque = 0

The Attempt at a Solution

Am i doing this correctly for part a? [/B]

 

[Jilang]

Are we to assume that there are no forces on the bolts?

 

[SteamKing]

You are introducing an extra quantity d₁ into your calculations, which will be difficult to calculate. Find an expression for sin θ based on L and H for the gantry crane as shown in the diagram.

 

[Aristotle]

You are introducing an extra quantity d₁ into your calculations, which will be difficult to calculate. Find an expression for sin θ based on L and H for the gantry crane as shown in the diagram.

Either way even with replacing d1 with H, I still get Ft = (mgx+MgL/2) /(L sin(theta) ) . Am I on the right step?

 

[SteamKing]

Either way even with replacing d1 with H, I still get Ft = (mgx+MgL/2) /(L sin(theta) ) . Am I on the right step?

d₁ ≠ H, and the expression for sin θ is a little more complicated than that. Your expression for F_T is correct.

 

[Aristotle]

d₁ ≠ H, and the expression for sin θ is a little more complicated than that. Your expression for F_T is correct.

If I can't call the line of the pivot point perpendicular to the cable d1, what should I call it? I'm a little confused on what you mean. Isn't that all I'm looking for anyways: the tension in the cable as a function of the load position x? - which was the expression i found?. :/

 

[Jilang]

Yes, it's absolutely fine. But can you express sin theta in terms of H and L ?

 

[Aristotle]

Yes, it's absolutely fine. But can you express sin theta in terms of H and L ?

Yes you would get sin(theta) = H/L ----> H=Lsin(theta). So then my final answer would be Ft = (mgx+MgL/2) /(L sin(theta) ) ----> Ft = (mgx+MgL/2) /(H). Much better I hope? :]

 

[SteamKing]

If I can't call the line of the pivot point perpendicular to the cable d1, what should I call it? I'm a little confused on what you mean. Isn't that all I'm looking for anyways: the tension in the cable as a function of the load position x? - which was the expression i found?. :/

There's another triangle (which you are overlooking) which can give you an expression for sin θ. (Hint: the cable holding the support arm forms the hypotenuse of this triangle)

 

[Jilang]

Nope. Try again.

 

[Aristotle]

There's another triangle (which you are overlooking) which can give you an expression for sin θ. (Hint: the cable holding the support arm forms the hypotenuse of this triangle)

The only triangle I see is the top triangle and bottom triangle and the triangle on the left with d1.

sin theta = H/L

How else can you derive the sin theta?

 

[SteamKing]

The only triangle I see is the top triangle and bottom triangle and the triangle on the left with d1.

sin theta = H/L

How else can you derive the sin theta?

sin θ is not H / L. You should review your trigonometry, especially the definitions of the sine, the cosine, and the tangent.

 

[Aristotle]

sin θ is not H / L. You should review your trigonometry, especially the definitions of the sine, the cosine, and the tangent.

Sorry about that, sin theta = H / sqrt (L^2 + H^2) :D