# Piano, Crane, and Two Ropes

1. Oct 20, 2007

### 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.

2. Oct 20, 2007

### Staff: Mentor

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?

3. Oct 20, 2007

### 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.

4. Oct 20, 2007

### Staff: Mentor

Right.

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

Show how you got that.

5. Oct 20, 2007

### 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

Thanks

6. Oct 20, 2007

### Staff: Mentor

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

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.

7. Oct 20, 2007

### 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??

8. Oct 20, 2007

### Staff: Mentor

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

9. Oct 20, 2007

### TonkaQD4

Doesn't that seem like an awful lot.

10. Oct 20, 2007

### Staff: Mentor

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