How to setup linear force equations

[Superman]

A planter of mass M=3kg is hung from a rod of mass m=1kg (3/4) of the way to the end. There is a horizontal wire attached to end of the rod pulling on it with tension Ft. The botom of the rod is hinged to the wall making an angle of 60 degrees with the wall. The assembly is to remain at rest.

1. Setup the linear force balance equations, one along each axis you choose.
2. Choose a pivot point and setup the torque balance about that point.
3. Solve for the tension in the wire and the two force components at the hinge.

I know I am just blatantly asking a question but this is the final question on my homework and I do not have a single clue how to do it. If you could explain all the steps involved please.

for the first one I tried think of it like, going down but I don't know how to get that into equations

I really need some help on this

 

[SteamKing]

Have you tried making a sketch of the planter, the rod, and the wire?

 

[Superman]

Have you tried making a sketch of the planter, the rod, and the wire?

Yes I made the sketch and I believe I have solved number 2

0=T[Lcos(theta)]-Mg[(3/4L)sin(theta)]-mg[(L/2)sin(theta)]

I still don't get number 1, do you know how to start number 1 once you have the sketch?

 

[SteamKing]

Did you show the forces acting on the rod on your drawing? You are given the mass of the rod and the planter, and the tension force in the wire is to be called Ft.

I don't understand how you could solve 2) without doing anything about 1).

 

[rezeew]

To setup the linear force equations, we need to consider the forces acting on the system and apply Newton's Second Law (F=ma) to each axis. In this case, we can choose the x-axis to be parallel to the wall and the y-axis to be perpendicular to the wall.

On the x-axis, we have the tension force Ft acting in the positive direction, and the component of the weight of the planter and rod system, Wx, acting in the negative direction. The equation would be:

Ft - Wx = 0

On the y-axis, we have the normal force, N, acting in the positive direction, and the remaining component of the weight, Wy, acting in the negative direction. The equation would be:

N - Wy = 0

To find the tension in the wire, we need to choose a pivot point and set up the torque balance equation. We can choose the hinge point as the pivot point, and the torque equation would be:

T = (m+M)gcos(60°) - Fty

Where T is the torque, m and M are the masses of the rod and planter respectively, g is the acceleration due to gravity, and Fty is the vertical component of the tension force.

Since the system is at rest, the net torque must be equal to zero. Therefore, we can solve for Fty:

Fty = (m+M)gcos(60°)

To find the two force components at the hinge, we can use the equations for the x and y components of the tension force:

Ftx = Ftcos(60°)
Fty = Ftsin(60°)

Substituting the value of Fty from the torque equation, we get:

Ftx = (m+M)gcos²(60°)
Fty = (m+M)gsin(60°)cos(60°)

Solving these equations will give us the values of Ftx and Fty. We can then use these values to find the normal force, N, by using the y-axis equation:

N = Wy = (m+M)gsin(60°)

I hope this helps you understand the steps involved in setting up linear force equations for this problem. It is important to carefully consider the forces acting on the system and choose appropriate axes and pivot points to accurately solve the problem.