Horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothesline, it sags a distance of 1 meters. What is the magnitude of the tension on the ends of the clothesline

T = 104N

Explanation:

We are to calculate the magnitude of the tension on the ends of the clothesline and to solve this, we need to draw a free body diagram to make it clearer.

Thus, i have attached the free body diagram.

Now, from the diagram, the angle made by the rope with the pole is given as α.

Now, we see that the rope sags by 1m. Since the weight is at midpoint, the point from the centre to the pole is 14/2 = 7m

Thus,using formula for finding angles in a right angle triangle, we have;

Tan α = opposite/adjacent = 1/7 =0.1429

Thus, α = tan^(-1)0.1429 = 8.133°

Now, to find the tension on the ends of the cloth line, we have to resolve the forces vertically.

Thus, still using the formula for finding angles in a right angle triangle, we have;

T sinα + T sinα = mg

i.e sum of upward forces = sum of dowward forces.

Thus,

2T(sin 8.133°) = 3 x 9.81

T(2 x 0.1415) = 29.43

T = 29.43/0.283 = 104 N

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y