Tension in a rope with weight held by a muscle builder

[Adi1973]

Homework Statement

A muscle builder holds the end of a massless rope. At the end of the rope, a 15 kg ball is hung as shown in the figure. What is the tension in the rope if the angle $$\theta$$ in the drawing is 4.5$$^{}o$$?

Homework Equations

g = 9.8m/s$$^{}2$$
F = mg
Trigonometrical functions

The Attempt at a Solution

I attached a picture. My answer is just totally wrong... but I have no idea what I'm doing wrong. Please could you help out. I apologise for any stupid mistakes!

The ball is being pulled by the two hands, forming a triangle that can be divided into two right triangles with sides labeled a1, b1, c1 and a2, b2, c2 respectively.

θ Is given as 4.5°

b1 = b2 = mg = 9.8N x 15N (weight of ball given as 15kg = 15N, gravitational pull is 9.8m/s2 = 9.8N)
= 147N

The tension of the rope would equal the force created by the hands pulling (a1 and a2) and a1 = a2.

sinθ = a1/b1
∴ a1 = b1 sinθ
= (147N)sin4.5°
= 12N

 

[tiny-tim]

Welcome to PF!

Hi Adi1973! Welcome to PF!

(try using the X² and X₂ icons just above the Reply box )

b1 = b2 = mg = 9.8N x 15N (weight of ball given as 15kg = 15N, gravitational pull is 9.8m/s2 = 9.8N)

sinθ = a1/b1

noooo … sinθ = b₁/a₁

sin = opp/hyp, cos= adj/hyp, tan = opp/adj

(and 15kg is not 15N, it's 15kg !

similarly, 9.8m/s2 is not 9.8N, it's … ? )

 

[Adi1973]

Hi tiny-tim, thank you so much for replying. I knew I was going to make someone cry with my answer...

Firstly, after having consulted my subject material, thanks I found the correct formula with regards to gravity, weight and work done: m/s2 x kg = N and yes, it doesn't mean I can just "convert" the m/s2 and kg values both to Newton... sorry.

Secondly, sinθ = b1/a1 - yes! It is actually one of the things I thought I finally got memorised but it's early days, so I'm obviously still making mistakes while turning the triangle in my head to find the right this over that formula. Thanks for pointing that out.

So, new answer (keep the tissues close):

N = gm
= 9.8m/s2 x 15kg (weight of ball given as 15kg, gravitational pull is 9.8m/s2)
= 147N

θ Is given as 4.5°
sinθ = b1/a1
∴ a1 = b1 /sinθ
= 147N/sin4.5°
= 1900N (1873.5 rounded to two significant figures)

Is that right?

 

[tiny-tim]

Hi Adi1973!

So, new answer (keep the tissues close):

N = gm
= 9.8m/s2 x 15kg (weight of ball given as 15kg, gravitational pull is 9.8m/s2)
= 147N

θ Is given as 4.5°
sinθ = b1/a1
∴ a1 = b1 /sinθ
= 147N/sin4.5°
= 1900N (1873.5 rounded to two significant figures)

Is that right?

ohh … so close …

only 100% out! ​

 

[Adi1973]

Do you have to divide the answer by two to get the tension, as it is shared on two sides (two hands pulling?)

 

[tiny-tim]

Hi Adi1973!

Do you have to divide the answer by two to get the tension, as it is shared on two sides (two hands pulling?)

hmm …yees, but … you're not thinking or talking like a physicist yet …

draw the free body diagram …

then say
"There are three forces: the weight downwards, and two equal tension forces

The horizontal components must add to zero, so they must be equal, and the vertical components must add to zero, so mg = … "

 

[Adi1973]

Do you mean that the downward force should be negative and the upward one positive? So mg = -147N? I'm sorry, I think I'm getting more lost now!

 

[tiny-tim]

Hi Adi1973!

Do you mean that the downward force should be negative and the upward one positive?

If you're measuring both the same way, then yes.

If you're measuring everything upward, then the weight is negative and the component of the tension is positive,

if you're measuring everything downward, then the weight is positive and the component of the tension is negative …

in either case, because the system is in equilibrium, everything adds to zero.