Help with Beam Physics Problem - Get the Answer Now!

  • Thread starter Thread starter SsUeSbIaEs
  • Start date Start date
  • Tags Tags
    Beam Physics
AI Thread Summary
A 9m beam with a mass of 300 kg is positioned over a ledge, and a 75 kg student intends to walk to its end. To determine how far the beam can extend from the ledge, the problem requires applying the condition for rotational equilibrium, where the sum of torques must equal zero. The equation set up involves the torques from both the beam's weight and the student's weight. Solving the equation reveals that the beam can safely extend 2.25m from the edge of the ledge. This solution ensures the beam remains stable while the student walks on it.
SsUeSbIaEs
Messages
17
Reaction score
0
This is probably a real easy problem but I can not get this problem right for my life, so would someone please help me out here :redface:


A 9m beam of mass 300 kg extends over a ledge, the beam is not attatched, but simply rests on the surface. A 75 kg student intends to position the beam so that he can walk to the end of it. Hwo far from the edge of the ledge can the beam extend?
 
Last edited:
Physics news on Phys.org
The sum of the moments needs to be zero or, assuming the beam extends out to the right, positive counterclockwise.

Can you post what you've tried so far?
 



Sure, I'd be happy to help you with this beam physics problem! First, let's start by drawing a free-body diagram of the beam and the student. We have the weight of the beam acting downwards at its center of mass, and the weight of the student acting downwards at the end of the beam. We also have a normal force acting upwards at the end of the beam where it rests on the surface.

Next, we can apply the condition for rotational equilibrium, which states that the sum of the torques acting on an object must be equal to zero. In this case, we have two torques - one from the weight of the beam and one from the weight of the student.

To solve for the distance from the edge of the ledge, we can set up the following equation:

Torque from beam weight = Torque from student weight

We know that the torque from the beam weight is equal to the weight of the beam multiplied by the distance from the center of mass to the end of the beam. Similarly, the torque from the student weight is equal to the weight of the student multiplied by the distance from the end of the beam to the edge of the ledge.

So, we can set up the equation as follows:

(300 kg)(9.8 m/s^2)(x) = (75 kg)(9.8 m/s^2)(9m - x)

Where x represents the distance from the end of the beam to the edge of the ledge. Solving for x, we get x = 2.25m.

Therefore, the beam can extend 2.25m from the edge of the ledge for the student to safely walk on it. I hope this helps! Let me know if you have any other questions.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top