How Do You Calculate the Tension in a Rope Holding a Leaning Beam?

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To calculate the tension in the rope holding a leaning beam, it is essential to consider the forces acting on the beam, including gravitational force, normal forces, and friction. The beam's weight creates torque around its base, and the tension in the rope also contributes to the torque balance. Setting the net torque to zero is crucial for equilibrium, but it is equally important to account for net forces acting in both horizontal and vertical directions. The equations provided need clarification regarding the forces' directions and their points of application. Understanding these dynamics is key to solving for the tension in the rope effectively.
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Homework Statement


A 24 kg beam of length 2.4 m leans against a smooth wall. A horizontal rope is tied to the wall and holds the beam on a frictionless floor. The beam makes an angle of 55 degrees with the floor. WHat is the tension in the rope? (The rope is on the floor.


Homework Equations


torque=force|| x distance


The Attempt at a Solution


(Torque at the bottom of the beam)
24kg x 9.8 x cos55 x 1.2m + T x cos35 x 2.4m = F(normal1) x cos55 x 2.4m

(torque at the top)
24kg x 9.8 x cos55 x 1.2m = F(normal2) x cos35 x 2.4m + Ff x cos55 x 2.4m

Now what?
 
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I don't understand your equations. Please identify all the forces acting on the beam: where they act, their direction and symbol.

Also realize that setting net torque equal to zero is only one of the conditions for equilibrium. What about the net force?
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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