Maximum tension in a cable holding a drawbridge

[greg_rack]

Homework Statement

A drawbridge system consists of a uniform ramp, of weight W , that is smoothly hinged at its lower end. The upper end is connected by a light, inextensible cable to a winch that is fixed to the wall in the position shown in the diagram(image below).
The ramp is lowered slowly, at constant speed, from its closed (vertical) position (θ = 0°) to its open (horizontal) position (θ = 90°).
What is the maximum tension in the cable during this process? (double-angle identities: sin 2θ = 2sin θ cos θ ; cos 2θ = cos2 θ – sin2 θ )

Relevant Equations

Weight
Trigonometry

My result for this problem is correct, but I would like to submit to you the method I used since it doesn't convince me.
The problem asks for the maximum tension on the cable from the transition to 0 to 90 degrees; I firstly concluded just by thinking, not using formulas or particularly formal methods, that the maximum tension must have been exerted for Θ=90°. Consequently, I started writing down the free
body diagram, including the weight applied to the middle of the drawbridge and the cable tension applied to the right corner of the drawbridge.
Now, placing the total momentum Mtot=0, I found the final formula and the correct answer... but I have a question: why wouldn't it have worked by placing just the total forces Ftot=0? Is that a non-working condition in problems consisting of a rotating body around a pivot?
I'm sorry but its a lot of time that I'm not studying such topics... and sometimes I Get confused!

 

[TSny]

My result for this problem is correct, but I would like to submit to you the method I used since it doesn't convince me.
... I firstly concluded just by thinking, not using formulas or particularly formal methods, that the maximum tension must have been exerted for Θ=90°.

Can you give us your particular thought process that led you to conclude that the maximum tension occurs at 90^o?

The fact that they gave you some trig identities makes me think that you are expected to derive an expression for the tension for any angle θ.

... but I have a question: why wouldn't it have worked by placing just the total forces Ftot=0? Is that a non-working condition in problems consisting of a rotating body around a pivot?

F_tot = 0 is a valid formula for this problem. But, the tension in the cable and the gravitational force are not the only forces acting on the bridge. The hinge also exerts a force on the bridge. This force would have to be included in F_tot.

Can you see why you didn't need to include the force at the hinge in M_tot?

 

[greg_rack]

Can you give us your particular thinking process that led you to conclude that the maximum tension occurs at 90^o?

Since the vertical weight component is maximized at 90°, and since its the factor which leads to an increase in the tension on the cable, I have immediately deducted it must have been maximum for θ=90°... but that doesn't convince me, its too empirical!

The fact that they gave you some trig identities makes me think that you are expected to derive an expression for the tension for any angle θ.

Exactly... but how? Deriving a general formula for any angle θ looks like quite tedious...

Can you see why you didn't need to include the force at the hinge in Mtot?

Yup, I got the point! Of course the force exerted by/on the hinge isn't relevant speaking in terms of momentum, since any force applied to the pivot has a related M=0.

 

[TSny]

Since the vertical weight component is maximized at 90°, and since its the factor which leads to an increase in the tension on the cable, I have immediately deducted it must have been maximum for θ=90°... but that doesn't convince me, its too empirical!

The vertical weight component is always mg. So it doesn't change. But, if you mean that the moment (about the hinge) due to the weight is maximized at 90^o, you are right. But that, by itself, is not enough to conclude that the tension in the cable must be a maximum at 90^o. It just means that the moment (about the hinge) due to the tension is maximized. The moment due to a force depends not only on the magnitude of the force, but also the direction of the force.

Exactly... but how? Deriving a general formula for any angle θ looks like quite tedious...

The mantra of the mechanics portion of any introductory physics course is "free-body diagram".

Yup, I got the point! Of course the force exerted by/on the hinge isn't relevant speaking in terms of momentum, since any force applied to the pivot has a related M=0.

Yes. If you take the hinge as the origin for calculating moments, then the force at the hinge has zero moment. (I know you meant "moment", rather than "momentum".)

 

[greg_rack]

The vertical weight component is always mg. So it doesn't change. But, if you mean that the moment (about the hinge) due to the weight is maximized at 90^o, you are right. But that, by itself, is not enough to conclude that the tension in the cable must be a maximum at 90^o. It just means that the moment (about the hinge) due to the tension is maximized. The moment due to a force depends not only on the magnitude of the force, but also the direction of the force.

I have managed to generalize the procedure by getting Tension depending on Weight and sin(θ/2), concluding it must have been maximized for the greatest theta possible, which in this case is indeed 90°!

 

[Lnewqban]

I have managed to generalize the procedure by getting Tension depending on Weight and sin(θ/2), ...

Do you mind showing us your work?