Question

A 0.13 kg meter stick is held perpendicular to a vertical wall by a 2.6 m string going from the wall to the far end of the stick.

A. Find the tension of the string

B. Find the tension in a 2.0 string.

Answer 1

A) There are two forces acting on the meter stick. The tension force by the string and the force due to weight. These two forces must balance each other so that the meter stick remains in static equillibrium

The horizontal force is

?F_x = N – Tcos? = 0

where,

Tcos? is the horizontal component of tension

N is the normal force.

implies that

N = Tcos? (1)

The vertical forces

?F_y = Tsin? – mg = 0

Tsin? is the vertical component of tension,

-mg is the weight of meter stick acting downwards.

we thus have

mg = Tsin? (2)


Dividing these two equations, we get

mg / N = Tsin? / Tcos?
or N = mg / tan? (3)

thus eq. (1) becomes

mg / tan? = Tcos?
T = mg / (cos?*tan?) (4)

The angle can be found as:

cos? = 1.0m / 2.6m
? = cos?¹(1.0m / 2.6m)
= 67.38°

Therefore eq. (4) becomes

T = (0.13 kg)(9.8 m/s²) / (cos67.38° x tan67.38°)
= 1.38 N

B) for the second part, same equations will be used, the angle will change and the tension accordingly.

T = mg / (cos?*tan?)   

cos? = 1.0m / 2.0m
? = cos?¹(1.0m / 2.0m)
= 60°

therefore,

T = (0.13 kg)(9.8 m/s²) / (cos60° x tan60°)
= 1.47 N