Calculate the force on the other support

[H_Ab]

Homework Statement

A uniform thin rod of weight 100 N is supported horizontally by two bricks at its
ends. At t=0 one of these bricks is kicked out quickly. Calculate the force on the
other support immediately thereafter.

Homework Equations

I'm not sure if I am on the right track with this equation since I don't have the length and from what I know, I need the length for torque problems: Mg(L/2)sin(theta)

The Attempt at a Solution

I thought since each brick has a force of 50N that if one brick is locked the force on the supporting brick is -50N since it's a downward force. Please help me with the equation and how to go about it. Thanks

 

[haruspex]

I'm not sure if I am on the right track with this equation since I don't have the length and from what I know, I need the length for torque problems: Mg(L/2)sin(theta)

Follow that through and see what happens. Maybe the length will not feature in the answer.
(From dimensional analysis, you can predict that it won't. Given an acceleration (g) a mass and a length as input, and a force to be the result of multiplying and dividing these and raising them to powers, the only combination that works is acceleration * mass * constant.)

 

[H_Ab]

So what would my constant be? My weight would be 50N since it's for one brick, and my acceleration is 9.81m/s. I'm confused about the constant.

 

[haruspex]

So what would my constant be?

You need to determine that by working through the usual equations.
Create unknowns for mass, length, angular acceleration and write out the standard equations for linear acceleration versus force and angular acceleration versus torque.

 

[HalfLostAndSearching]

I would approach this problem by first drawing a free body diagram of the rod and the forces acting on it. The forces acting on the rod are its weight (100 N) acting downwards, and the reaction force from the two supports acting upwards. Since one of the supports is kicked out, the reaction force from that support will no longer be present. This means that the only force acting on the remaining support is the weight of the rod (100 N).

Using Newton's second law, F=ma, we can calculate the acceleration of the rod immediately after the support is kicked out. Since the rod is initially at rest and there is no net force acting on it, the acceleration will be 0 m/s^2.

Next, we can use the equation F=ma again to calculate the reaction force on the remaining support. Since the acceleration is 0 m/s^2, the reaction force must also be 0 N. This means that there is no force acting on the remaining support immediately after the other support is kicked out.

In summary, the force on the other support immediately after the support is kicked out is 0 N. This makes sense intuitively, as the rod is no longer being supported and will start to fall due to its weight.