What Are the Tensions Required to Steady a Piano Being Lowered by a Crane?

[TonkaQD4]

A 500kg piano is being lowered into position by a crane while two people steady it with ropes pulling to the sides. Bob's rope pulls to the left, 15 degrees below the horizontal, with 500 Newtons of tension. Ellen's rope pulls toward the right, 25 degrees below the horizontal.

(a) What tension must Ellen maintain in her rope to keep the piano desceding at a steady speed?

(b) What is the tension in the main cable supporting the piano?

Well of course I started out by drawing the situation. Bob's rope in the third quadrant at 15 degrees and Ellen's rope in the fourth quadrant at 25 degrees below the x-axis.

 

[Doc Al]

Realize that Bob and Ellen are merely steadying the crane, so the crane cable supporting the piano remains vertical. What must be true about the sum of the horizontal forces on the piano?

 

[TonkaQD4]

The horizontal forces are Zero

Due to the complexity and length of this problem I was wondering if anybody else could complete it and then actually compare answers, and then if we differ, then maybe we could go into greater detail.

Basically using trig I found Bob in terms of Ellen and came up with an answer of...
In order to keep the piano steady Ellen must apply Tension to her rope of 724.34N.

 

[Doc Al]

The horizontal forces are Zero

Right.

Due to the complexity and length of this problem I was wondering if anybody else could complete it and then actually compare answers, and then if we differ, then maybe we could go into greater detail.

Only one equation is needed: That horizontal forces add to zero.

Basically using trig I found Bob in terms of Ellen and came up with an answer of...
In order to keep the piano steady Ellen must apply Tension to her rope of 724.34N.

Show how you got that.

 

[TonkaQD4]

Ok
I made Bob's vector F_B and Ellen's vector F_E

First I solved for the x terms...

F_x=ma_x

-F_Bcos15+F_Ecos25 = 0 so,
F_E = F_B(cos15/cos25)

Now the y terms...

F_y - mg = ma_y

-F_Bsin15+(-F_Esin25)-mg = ma_y
-F_Bsin15-F_B(cos15/cos25)(sin25)= mg

-mg = F_Bsin15 + F_Bcos15tan25


Therefore...

F_E = F_B (cos15/cos25)

= -mg / (sin15+cos15tan25) (cos15/cos25)

= -mg / (cos25/tan15) + (cos25/tan15)

= -mg / 2cos25/tan15

Conclusion F_E

F_E = -500(9.8) / (2cos25/tan15)

= 4900 / 6.76

724.34 NEWTONS



I think I might be making this harder than it really is, please give me your best advice, but I might have to ask my teacher about this one.

Thanks

 

[Doc Al]

Ok
I made Bob's vector F_B and Ellen's vector F_E

First I solved for the x terms...

F_x=ma_x

-F_Bcos15+F_Ecos25 = 0 so,
F_E = F_B(cos15/cos25)

This is perfectly correct and is all you need for part (a)! You are given that F_B = 500, so just plug it into find F_E.

Now the y terms...

F_y - mg = ma_y

-F_Bsin15+(-F_Esin25)-mg = ma_y
-F_Bsin15-F_B(cos15/cos25)(sin25)= mg

-mg = F_Bsin15 + F_Bcos15tan25

Unfortunately, this is not correct because you did not take into account the force that the cable exerts on the piano. Since the piano is not accelerating, the vertical forces must also equal zero. Combined with the other equation you can figure out the cable tension--that's what you need to do to solve part (b).

Rewrite your equation for the y forces, this time including the upward force of the cable tension.

 

[TonkaQD4]

Well
mg
500kg(9.8m/s^2)

=4900N

Well.. I'm not exactly sure how to solve this ...

-F_Bsin15 - F_Esin25 - mg + F_t = ma = 0

F_t = F_Bsin15 + F_Esin25 + mg

= 129.41N + 225.21N + 4900N

F_t = 5254.62N ----> this can't be correct, what am I doing wrong??

 

[Doc Al]

Looks good to me. (Why do you say it can't be correct?)

 

[TonkaQD4]

Doesn't that seem like an awful lot.

 

[Doc Al]

Doesn't that seem like an awful lot.

The piano's heavy! It weighs 4900 N. And you have two people tugging down on it.