Finding tension of 2 wires at different angles

[rollad]

A bag of cement of weight 325N hangs from 3 wires as shown in figure (I've attached it). Two of the wires make angle₁=60^oand angle₂=25^o with the horizontal. If the system is in equilibrium, determine the tensions of T1, T2, and T3

If you can't see the image, the T1 and T2 are suspended from a horizontal at the angles specified. T1 on the left, T2 on the right. Where they come together T3 is hanging which is attached to the cement with weight 325N. I hope this is helpful.

So I am assuming that T3=325N
I set up the net forces in the y direction as 0=T1sin60+T2sin25-325
That is where I am stuck. I don't understand how to find the tensions from this. I thought about using the forces in the x direction to solve for either T1 or T2 and then substituting that in for one of the other, however that just does not make sense to me. This is my second semester in physics and the first time I took this class this was one of my biggest problems, I just don't know what to do when I get to this point. Thanks for any and all help, I surely appreciate it.

 

[TSny]

Hi, rollad.

So I am assuming that T3=325N
I set up the net forces in the y direction as 0=T1sin60+T2sin25-325

Looks good.

That is where I am stuck. I don't understand how to find the tensions from this. I thought about using the forces in the x direction to solve for either T1 or T2 and then substituting that in for one of the other, however that just does not make sense to me.

Actually, that makes a lot of sense to me! What bothers you about it?

 

[rollad]

This is what I get 0=(T2cos25/cos60)sin60+T2sin25-325

I guess I am just confused on how to get everything equally a single T2

 

[TSny]

This is what I get 0=(T2cos25/cos60)sin60+T2sin25-325

I guess I am just confused on how to get everything equally a single T2

OK. That looks good. The rest is just simplifying the expression. The first two terms on the right can be combined into one term involving T2. Recall that $ax + bx = (a+b)x$. You can boil all those trig functions down to a single number.

 

[Nickie]

I would approach this problem by first identifying all the known and unknown variables. In this case, we have the weight of the cement (325N) and the angles of the wires (60o and 25o) as known variables, and the tensions (T1, T2, T3) as unknown variables.

To solve for the tensions, we can use the principle of equilibrium, which states that the net force and net torque acting on an object must be equal to zero. In this case, we can set up equations for the net forces in the x and y directions.

In the x direction, we have T1cos60 + T2cos25 = 0, since the weight of the cement does not have a component in the x direction. We can rearrange this equation to solve for T1 or T2, depending on which variable we want to solve for first.

In the y direction, we have T1sin60 + T2sin25 - 325 = 0, since the weight of the cement is acting downwards and must be balanced by the upward tension forces. We can solve for either T1 or T2 using the equation we found in the x direction, and then substitute that value into this equation to solve for the other tension.

Once we have solved for T1 and T2, we can use the principle of equilibrium again to solve for T3. The net torque acting on the cement must also be equal to zero, so we can set up an equation for the torques in this system. We can use the known angles and the tensions we have already solved for to find the torque acting on the cement, and then set it equal to zero.

In summary, to find the tensions of the wires at different angles, we can use the principles of equilibrium and set up equations for the net forces and torques acting on the system. By solving these equations, we can find the tensions of the wires and ensure that the system is in equilibrium.