A wet towel of mass 4 kg is hung in the center of a 6 m clothesline, so that the line makes an angle 10o with the horizontal. The tension in the line is closest to...
50N, 40N, 100N, 200N, or 10N
each half of the closline supports 2kg
draw the triangle:
sin10=2g/tension
tension=about 20/sin10=20/(10/57) using the small angle approximation
tension=114/10= about 10 N
To find the tension in the clothesline, we can start by visualizing the forces acting on the towel. When the towel is hung, the weight of the towel acts downwards, and the tension in the line acts upwards and at an angle of 10 degrees with the horizontal.
First, we can break down the weight of the towel into its horizontal and vertical components. The vertical component of the weight would be mg, where m is the mass of the towel (4 kg) and g is the acceleration due to gravity (9.8 m/s^2). Therefore, the vertical component of the weight is 4 kg * 9.8 m/s^2 = 39.2 N.
Next, we can find the horizontal component of the weight by multiplying the vertical component by the tangent of the angle. In this case, the angle is 10 degrees, so the tangent would be tan(10) = 0.176. Therefore, the horizontal component of the weight is 0.176 * 39.2 N = 6.9 N.
Since the tension in the line needs to balance both the vertical and horizontal components of the weight, we can find the tension by calculating the resultant force. The resultant force can be found using the Pythagorean theorem:
Tension^2 = (horizontal component of weight)^2 + (vertical component of weight)^2
Plugging in the values:
Tension^2 = (6.9 N)^2 + (39.2 N)^2
Tension^2 = 47.61 N^2 + 1,536.64 N^2
Tension^2 = 1,584.25 N^2
Taking the square root of both sides, we find:
Tension ≈ 39.8 N
Therefore, the tension in the line is closest to 40N.