Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspend
ID: 2894674 • Letter: R
Question
Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 10 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension. (Use g = 9.8 m/s2 for the acceleration due to gravity. Round your answers to two decimal places.) Really just need the tensions, not the magnitudes
3 m rope tension:
3 m rope magnitude:
5 m rope tension: and 5 m rope magnitude:
Explanation / Answer
Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 10 kg. The ropes, fastened at different heights, make angles of 52° (3m rope) and 40° (5m rope) with the horizontal. Give your answers correct to two decimal places.
When I do this type of vector problem, I draw a scale drawing of 2 upside down right triangles!!
1st upside down right triangle
The hypotenuse of the right triangle represents the 3 m rope.
In the scale drawing, 3 cm = 1 m.
Draw a 6 cm horizontal line.
Label the left end as Point A.
Draw a 9 cm line from Point A at an angle of 52° below horizontal and to the right of Point A.
Label the bottom right end of this line as Point B.
Draw a vertical line from Point B up to the horizontal line.
Label the point where this vertical line intersects the horizontal line as Point C.
You should see a right triangle, ACB, with the 52° angle at Point A and the right angle at Point C.
Line AB is the hypotenuse of the right triangle.
Angle CAB is 52°.
Line AB represents the tension in the 3 m rope.
Line CB is the vertical component of the tension.
Line AC is the horizontal component of the tension
Tension in 3 m rope = T3
Vertical component of the tension in 3 m rope = T3 * sin 52°
Horizontal component of the tension in 3 m rope = T3 * cos 52°
2nd upside down right triangle
The hypotenuse of the right triangle represents the 5 m rope.
In the scale drawing, 2 cm = 1 m.
Draw a 8 cm horizontal line.
Label the right end as Point A.
Draw a 10 cm line from Point A at an angle of 40° below horizontal and to the left of Point A.
Label the bottom left end of this line as Point B.
Draw a vertical line from Point B up to the horizontal line.
Label the point where this vertical line intersects the horizontal line as Point C.
You should see a right triangle, ACB, with the 52° angle at Point A and the right angle at Point C.
Line AB is the hypotenuse of the right triangle.
Angle CAB is 40°.
Line AB represents the tension in the 5 m rope.
Line CB is the vertical component of the tension.
Line AC is the horizontal component of the tension
Tension in 5 m rope = T5
Vertical component of the tension in 5 m rope = T5 * sin 40°
Horizontal component of the tension in 5 m rope = T5 * cos 40°
The 2 vertical components support the weight of the decoration.
Weight = 10* 9.8 = 98 N
Equation #1
T3 * sin 52° + T5 * sin 40° = 98
The 2 horizontal components prevent the decoration from accelerating to the left or right.
Equation #2
T3 * cos 52° = T5 * cos 40°
We have 2 equations with 2 variables, solve for T3 in Equation #2, and then substitute into Equation #1.
Equation #2
T3 * cos 52° = T5 * cos 40° = 0
Divide both sides by cos 52°
T3 = (T5 * cos 40°) ÷ cos 52°
T3 = T5 * (cos 40° ÷ cos 52)
substitute into Equation #1
Equation #1
T3 * sin 52° + T5 * sin 40° = 98
[T5 * (cos 40° ÷ cos 52)] * sin 52° + T5 * sin 40° = 98
T5 * [(cos 40° ÷ cos 52) * sin 52] + sin 40° = 98
T5 * 1.622 = 98
T5 = 60.42 N
Substitute into Equation #2
T3 * cos 52° = 60.42 * cos 40°
T3 = (60.42 * cos 40) ÷ cos 52
T3 = 75N
Find the tension in the 3 m rope.
146 i + 15.59 j
b) Find the magnitude of this tension.
T3 = 75N
a) Find the tension in the 5 m rope.
46.28 i + 38.8 j
b) Find the magnitude of this tension.
T5 = 60.42 N
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