Two strings each inclined at an angle 30o to the horizontal are used to hold a bucket of water weighing 20N. Calculate the tension i n each string.
20 (\sqrt{3}\)N
20N
10 (\sqrt{3}\)N
10N
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The correct answer is option B, 20N.
Since the angles given were to the horizontal, you convert to the vertical first by subtracting from 90°. That will give us an equilateral triangle of 60 degrees each.
Therefore 20/sin 60 = X/sin 60
X = 20x sin60/sin60
X =20 N
The free body diagram of the bucket shows that the weight of the bucket (20 N) is resolved into two components along each of the strings. Since the strings are inclined at an angle of 30 degrees to the horizontal, the angle between the weight vector and the strings is also 30 degrees. Therefore, each string has a tension component that balances the weight component along the string. The tension force T can be found by resolving the weight vector into components along the strings and using trigonometry.
The weight component along each string is:
Wsin(30) = 20N x sin(30) = 10 N
The tension force in each string is equal to the weight component along the string, so:
T = 10 N
Therefore, the correct answer is option D.
When dealing with strings or ropes at an angle, you can use trigonometry to find the tension in each string. In this case, the bucket of water weighs 20 N and the angle of inclination to the horizontal is 30 degrees.
Let's denote the tension in each string as T. Since the strings are inclined at the same angle, the tension will be the same in both strings.
The forces acting on the bucket of water are its weight (20 N) pulling downward and the tension in the strings pulling upward. These forces create a vertical equilibrium, meaning the net vertical force is zero.
The vertical component of the tension is given by T×sin(angle), where the angle is 30 degrees in this case.
T×sin(30 )=20N
T=20N / Sin 30
Using the sine of 30 degrees (which is1/2)
we can calculate
T:=20 N/1/2
T=40N
So, the tension in each string is 40 N.
