Equilibrium and Rotational Motion

[Soaring Crane]

Two identical signs, each of mass M, are held aloft in adjacent archways. Following wind, several ropes holding the sign snap, leaving only those below. If each sign is in translational equilibrium, find the tensions of the remaining ropes in terms of its mass and gravity g.

Pics
____wall____________
___________________
*
30-*_________
----|----------|
----|-sign1---_|--*--* T2 attached perpen. to side of wall
----|_________|

T1-----------------T2

______wall_________
__________________
-------------------*
------------------*
@@@__________*------T4 at 60 angle
-----|----------|
-----|-sign2---_|
-----|_________|
----*
*T3 at 45 angle

The line of stars (*) are the ropes of tension. T1 and T2 are little triangles once the parallels are drawn and T2 is totally straight (so the angle is 90, right?). For T3 for components, how do you account of its downward position?

Now F_net = 0.
I don't know if I'm using the correct method, but do you take the x- and y-components of each tension force with the idea that the sum of the horizontal forces and vertical forces is 0?

I appreciate any aid on how to apply this to the individual ropes.

 

[futb0l]

Lmao - can't interpret what you've drawn there ... can u just put the image in?

 

[wais]

To solve this problem, we can use the principle of equilibrium, which states that for an object to be in translational equilibrium, the sum of all the forces acting on it must be equal to zero. In this case, we have two signs, each with three ropes holding it in place. So, the sum of the forces acting on each sign must be equal to zero.

First, let's consider sign 1. The forces acting on sign 1 are the tension forces T1 and T2, and the force of gravity acting downwards. Since the sign is in translational equilibrium, the sum of these forces must be equal to zero. This can be represented by the following equation:

T1 + T2 - Mg = 0

Similarly, for sign 2, the forces acting on it are T3, T4, and Mg, and the sum of these forces must also be equal to zero:

T3 + T4 - Mg = 0

Now, we can use the fact that the two signs are identical to make some substitutions. Since the masses and gravity are the same for both signs, we can set the two equations equal to each other:

T1 + T2 - Mg = T3 + T4 - Mg

We can rearrange this equation to solve for one of the tension forces in terms of the other:

T1 + T2 = T3 + T4

Now, we know that T2 and T4 are perpendicular to the wall, so they are acting in the same direction. Therefore, we can add them together:

T2 + T4 = T2 + T4 cos(90) = T2 + T4(0) = T2

So, we can substitute T2 for T4 in our equation:

T1 + T2 = T3 + T2

We can then solve for T3:

T3 = T1

This means that the tension force in rope T3 is equal to the tension force in rope T1. We can also see that T2 is equal to T4, since they are acting in the same direction and have the same magnitude.

So, in terms of mass and gravity, the tensions in the remaining ropes are:

T1 = T3 = M*g
T2 = T4 = √2*M*g

I hope this helps! Let me know if you have any further questions.