Tension in a rope suspending a hinged plate

[Sulla]

Determine the tension in the support rope and the forces of the hinge on he uniform rectangular mass below. The length on the rect. is 12 and the height is 6m. The angle that the rope makes with the rectangle is 30 degrees.


____
I broke up the hinge forces into x and y components:

Bx=Tcos 30
By + Tsin 30= mg


mg(6) <--(half of the length) = T sin 30 (12)

T = mg(6)
--------- = mg
12(sin 30)

Now that I've calculated T, I use it to find out the forces of the hinge.

Bx = mg(cos 30)
By = (square root of) \[mg^2 - \[3/2] mg2]



That's as far as I got. I don't know how to determine the actual forces since there are no number provided for the mass. And I don't think that the Tension calculated is correct.

Also, does the space between the hinge and the object affect the tension in the rope? Are my calculations of the tension only determine it along the length of the object?


Any help would be much appreciated!

 

[PhanthomJay]

Determine the tension in the support rope and the forces of the hinge on he uniform rectangular mass below. The length on the rect. is 12 and the height is 6m. The angle that the rope makes with the rectangle is 30 degrees.


____
I broke up the hinge forces into x and y components:

Bx=Tcos 30
By + Tsin 30= mg


mg(6) <--(half of the length) = T sin 30 (12) +Tcos 30(6)[/color]

T = mg(6)
--------- = mg
12(sin 30)

Now that I've calculated T, I use it to find out the forces of the hinge.

Bx = mg(cos 30)
By = (square root of) \[mg^2 - \[3/2] mg2]what's this? By = mg - Tsin30[/color]



That's as far as I got. I don't know how to determine the actual forces since there are no number provided for the mass. Can't be determined without a value of m, so your answer must remain as a function of m.[/color] And I don't think that the Tension calculated is correct.see remarks above[/color]

Also, does the space between the hinge and the object affect the tension in the rope? Are my calculations of the tension only determine it along the length of the object?you have the correct free body diagram, the space does not matter[/color]


Any help would be much appreciated!

See comments in red above.

 

[Sulla]

So this portion:

T = mg(6)
--------- = mg
12(sin 30)

becomes
mg - T cos 30
-------------
T sin 30

after adding the T cos30 (6)?

Wouldn't that change the value of Bx and By though?

 

[Sulla]

There is also another part to this question that I need help with:

If the cable snaps, the rectangular mass will rotate about the hinge. Assume the rectangle begins to fall starting from the rest the instant the rope snaps and the moment of inertia about the center of mass, perpedicular to the page is I(bar)z = 450 Kgm^2

What is the moment of inertia of the rectangle about the hinge axis perpendicular to the page? and what will be the magnitude of the angular velocity of the rectangle as it swings to its lowest postion?I attempted to use the perpendicular axis theorum using the Bx and By forces of the pervious question, but since there are no actual numbers in those forces, I could not get a definate answer. Also, I could not set the I z to equal those of By and Bx since Bx and By are the forces of the hinge while Iz is the inertia of the center of mass of the rectangle.

Any suggestions as to how I can approach this problem?

Thanks!

 

[PhanthomJay]

So this portion:

T = mg(6)
--------- = mg
12(sin 30)

becomes
mg - T cos 30
-------------
T sin 30

after adding the T cos30 (6)?

Wouldn't that change the value of Bx and By though?

You're losing me with your algebra. It's
mg(6) = Tsin30(12) + Tcos30(6), now divide both sides of eq. by 6,
mg = 2Tsin30 + Tcos30
mg = T(2sin30 +cos30)
mg = T(1 + .866) = T(1.866)
solve T = mg/1.866

Then solve
Bx=Tcos 30 = (mg/1.866)(.866) = .464mg
and since
By + Tsin 30= mg, then
By = mg - Tsin30
By = mg - (mg/1.866)(1/2)
By = mg - .268mg
By = .732mg

 

[PhanthomJay]

There is also another part to this question that I need help with:

If the cable snaps, the rectangular mass will rotate about the hinge. Assume the rectangle begins to fall starting from the rest the instant the rope snaps and the moment of inertia about the center of mass, perpedicular to the page is I(bar)z = 450 Kgm^2

What is the moment of inertia of the rectangle about the hinge axis perpendicular to the page? and what will be the magnitude of the angular velocity of the rectangle as it swings to its lowest postion?


I attempted to use the perpendicular axis theorum using the Bx and By forces of the pervious question, but since there are no actual numbers in those forces, I could not get a definate answer. Also, I could not set the I z to equal those of By and Bx since Bx and By are the forces of the hinge while Iz is the inertia of the center of mass of the rectangle.

Any suggestions as to how I can approach this problem?

Thanks!

Are you sure the mass isn't given? You can calculate it by knowing the formula for the rotational moment of inertia of a rectangle about its c.m. Then use the parallel axis theorem to calculate I about the hinge, which is not a function of the hinge forces anyway. Are you familiar with the calcuations for I_cm and I_hinge? Once solved, you can calculate the angular speed at the bottom using conservation of energy prnciples.

 

[Sulla]

Okay, so I get:

I(cm) = 1/12M (L^2 + W ^2)
= 1/12M (12^2 + 6^2)
= 15M

The only other way I can think of to calcualte the speed is:
Torque = mg (L/2)
= m(9.8)(12/2)
T= 58.8M
T = I alpha
58.8M/15 = 15M/15
alpha = 3.92 rad/sec^2

Then I used an angular kinematic eq.
omega^2 = omega (initial) + 2(alpha)(theta initial - theta final)
omega^2 = 2(3.92) (pi/2) <----(assuming/hoping that the angle the rect. rotates is 90 degrees?)
omega = 3.51 rad/sec

I tried using cons. of energy but that didn't get me anywhere since the mass and H are not given.

 

[PhanthomJay]

Okay, so I get:

I(cm) = 1/12M (L^2 + W ^2)
= 1/12M (12^2 + 6^2)
= 15M Excellent! Now since it is given that I_cm = 450, you can easily solve for M = 30Kg . But now you must go one step further: since the sign will rotate about the hinge, you must calculate the I about the hinge using the parallel axis theorem: I_hinge = I_cm + MD^2, where D is the (diagonal) distance between the 2 axes.[/color]

The only other way I can think of to calcualte the speed is:
Torque = mg (L/2)this won't work , since Torque is not constant; T =mgL/2 in the horizontal position, but T=0 when the sign is at the bottom of its swing, and it is not a linear transition in between[/color]
= m(9.8)(12/2)
T= 58.8M
T = I alpha
58.8M/15 = 15M/15
alpha = 3.92 rad/sec^2no, T is not constant, as noted[/color]

Then I used an angular kinematic eq.
omega^2 = omega (initial) + 2(alpha)(theta initial - theta final)
omega^2 = 2(3.92) (pi/2) <----(assuming/hoping that the angle the rect. rotates is 90 degrees?)
omega = 3.51 rad/sec

I tried using cons. of energy but that didn't get me anywhere since the mass and H are not given. Try it again, you have M and H_cm is just L/2 at the lowest position[/color]

See comments above.