VECTORS!? PLEASE HELP!

Question

A clothesline is tied between two poles, 10m apart. The line is quite taut and has negligible sag. When a wet shirt with a weight of 0.8 kg is hung at the middle of the line, the midpoint is pulled down 10 cm. Find the tension in each half of the clothesline. (Use g = 9.8 m/s^2 for the acceleration due to gravity. Enter your answers from smallest to largest value of the coefficient of vector i.)

 

Vector T1 = ?

 

Vector T2 = ? 

 

PLEASE HELP! Thanks in advance! 

Answer 1

We are looking for two vectors--force vectors--one for each half of the clothesline.

 

These two vectors are the two forces that each half of the clothesline is exerting on the shirt.

 

These vectors point along the clothesline from the middle of the sag back up toward either support.

 

By symmetry, the vectors are going to be mirror images of each other, which means the x-components will cancel.

 

However, the y-components are going to add together to give a total force in the y-direction due to the clotheslines.

 

This is good because there needs to be a force to cancel gravity since the shirt is not accelerating (Fnet=m*a=0).

 

Indeed, we can use this to find the y-components of the two force vectors: they each have to be half the force due to gravity so that when they add up, the total cancels the force due to gravity:

 

F_y1=F_y2 (symmetry)

F_y1 + F_y2 = Fg (equilibrium)

=> 2F_y1=2F_y2=Fg

=> F_y1=F_y2=Fg/2

 

Since we know the force vectors must point along the clotheslines, we can use trigonometry and the dimensions of the clothesline combined with the depth of the sag to find the angles for the force vectors.

 

Then we can use trigonometry again to find the x-components of the force vectors using the angles and y-components.

 

 

 

Answer 2

By the symmetry of this problem, the magnitudes of the tensions are the same.  This may be hard to see without a diagram, but visualize two right triangles, of vertical side 10 cm = 0.1 m, and horizontal side 10/2 m = 5m.

 

Take the angle θ from the vertical:

 

θ = tan^-1(5/0.1) = 88.85°

 

Static equilibrium in the vertical direction:

 

2T cosθ = mg

 

T = mg / 2cosθ = (0.8 kg)(9.8 m/s²) / 2cos(88.85°) = 195.32 N

 

T_x = Tsinθ = (195.32) sin(88.85°) N = 195.28 N

 

T_y = Tcosθ = (195.32) cos(88.85°) N = 3.92 N

 

(Check that T_x² + T_y² = T²)

 

Tension on the right cable:

 

T_r = (195.28i + 3.92j) N

 

Tension on the left cable:

T_l = (-195.28i + 3.92j) N