Tension in Cable: Solving Using Trig

  • Thread starter Thread starter ChaosCon343
  • Start date Start date
  • Tags Tags
    Cable Tension
AI Thread Summary
To find the tension in the cable supporting a 36.0 kg beam at an angle of 69.0°, it's essential to consider the torque around the hinge. The initial attempt using basic trigonometry was ineffective due to the beam's center of mass not aligning with the cable's attachment point. A more effective approach involves calculating the torque about the hinge and applying equilibrium conditions. Utilizing the weight of the beam and the angle, one can derive the tension in the cable accurately. Proper application of torque principles will lead to the correct solution.
ChaosCon343
Messages
7
Reaction score
0

Homework Statement



A 36.0 kg uniform beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal.

http://a1.educog.com/res/msu/physicslib/msuphysicslib/24_Statics_Equilibrium/graphics/prob20_beamhinge2.gif

If the angle between the beam and the cable is θ = 69.0° what is the tension in the cable?

Homework Equations


Unknown other than basic trigonometry

The Attempt at a Solution


I tried using simple trigonometry; Cos(69)=(36*g)/h, but to no avail. I believe the reason it didn't work to be because the cable is not attached to the bar's center of mass, but I've no idea how to correct this. A point in the right direction would be greatly appreciated!
 
Last edited by a moderator:
Physics news on Phys.org
Use torque about hinge.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...

Similar threads

Back
Top