Thread: Cantilever beam: Solving for reaction forces

Would anyone be able to assist me in solving this puzzle? It should be simple, though I have no background in physics or math.

So here's the situation. Imagine you have a 5 meter long cantilever beam, with a 100kg man standing at 4.6498985 meters from the wall. A crane hook is attached to the end of the beam imparting an upward force on the beam of 936.92865 Newtons (N) (a Newton is a unit of force attained by multiplying a mass such as 1 kilogram by an acceleration such as the gravitational constant 9.81 meters per second per second here on Earth, i.e. 1kg X 9.81 m/s² = 9.81N). Solving for the reaction forces at the wall, keeping 6 or 7 decimal places in your calculation to avoid round off error.
I asked about this on a different forum, and just got a few very vague hints, one of which was definitely not right. I should explain briefly that while I appreciate hints/explanations, this is not for a school project or anything (I've been out of school for nearly two decades). I'm just looking for the answer to the problem, which is part of a puzzle I'm solving (for fun). So basically, all I need is a number.

Thank you in advance for any help you might have, I appreciate it.

It's possible that I am looking for two numbers, one around 44, and one around 123.

the beam is in equilibrium, yet the net torque about the point of the beam against the wall is unbalanced.

981(4.6498985) - 936.92865(5) = -123.0928215 Nm ... you have more clockwise than counter-clockwise torque

The additional ccw torque must come from the weight of the beam. Assuming the beam is uniform, it exerts a weight of 49.2371286 N downward at a distance of 2.5 m from the end of the beam.

also, since the beam is in equilibrium ...

forces up = forces down

936.92865 + R = 981 + 49.2371286

R = 93.3084786 N upward reaction force at the wall

Thank you, I really appreciate your help!

Hmm.. I ran in through the solution checker, and it didn't work, so I contacted the guy who made the puzzle. This is the explanation I got from him for why my number didn't work.

---

I'm not sure where you would have calculated the force 49.2371286. There are only two forces acting on the beam; the crane and me (technically the beam's mass is also acting on this situation but this is a more advanced problem). As such you should only have those two forces and the reaction force at the wall in your equation of equilibrium. Other than that you look good, when you have the correct answer it will be painfully obvious you are right or in the right ball park. Additionally your moment equation will only have two torques and the reaction moment at the wall as well.

---

I don't really get what he wants me to do, my own solution was 123.0928215 (man x distance - crane x distance), but that still isn't the answer.

if those are the only two torques about the wall end of the beam (due to the weight of the man and the upward force exerted by the crane hook) , then the system is not in rotational equilibrium as I first stated in my previous post