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# An iron bolt of mass 80.0 g hangs from a string 38.6 cm long. The top end of the string is fixed. Without?

touching it, a magnet attracts the bolt so that it remains stationary, displaced horizontally 23.0 cm to the right from the previously vertical line of the string. Find the tension in the string. Find the magnetic force on the bolt.

### 1 Answer

- electron1Lv 710 years agoFavorite Answer
An iron bolt of mass 80.0 g hangs from a string 38.6 cm long. The top end of the string is fixed. Without touching it, a magnet attracts the bolt so that it remains stationary, displaced horizontally 23.0 cm to the right from the previously vertical line of the string. Find the tension in the string. Find the magnetic force on the bolt.

The best way to determine what this problem really looks like is to draw scale drawing.

38.6 cm is longer than a ruler, so let’s draw a drawing that is ½ the size of the numbers in the problem

Use regular 8 ½ by 11 inch notebook paper.

Lay the paper like you usually do, so it is 8 ½ inches wide and 11.5 inches tall.

Darken the top line of the paper. This line represents the support beam that the top of the string is attached to.

Draw a 19.3 cm vertical line down the vertical red margin line from the top line of the paper. Label the top of this line as Point A, and the bottom as Point B. This vertical line represents the 38.6 cm string.

Measure 11.5 cm, on the top line of the paper, to the right of the point where the red margin line intersects the top line of the paper. Label the left end of this line as Point C.

Draw a 19.3 cm vertical line down from Point C. Label the bottom of this line as Point D. Draw line BD.

Set the 0 of the ruler on the Point A, You need to draw a 19.3 cm line from Point A to line CD. Just rotate the ruler until 19.3 cm intersects line CD. Label the point of intersection as Point E.

Draw a horizontal line from Point E to the left until it intersects line AB.

Label the point where this horizontal line intersects line AB as Point F.

Draw line FE.

You should see a rectangle ABDC, with a smaller rectangle AFEC as the top part of it.

Draw a line from Point A to Point E. This line, AE, represents the tension in the string, when the iron bolt has been displaced horizontally 23.0 cm to the right from the previously vertical line of the string.

The hypotenuse, line AE, of the right triangle AFE is 19.3 cm long.

The base, line FE, of the right triangle AFE is 11.5 cm wide.

The sine of Angle FAE = 11.5/19.3, so Angle FAE = 36.57º

Triangle FAE represents the vector drawing the weight of a 0.080 kg iron bolt that was hanging vertically from Point A, but now is pulled 23 cm horizontally to the right. The vertical line AF represents the weight of the 0.080 kg iron bolt.

The 0.080 kg iron bolt is hanging from the lower end of the line AB

The weight of the 0.80 kg iron bolt = 9.8 * 0.08 = 0.784 N

The hypotenuse, line AE, represents the tension in the string, when the iron bolt has been displaced horizontally 23.0 cm to the right from the previously vertical line of the string.

Line AF divided by Line AE = cosine of Angle FAE.

Line AE = Line AF ÷ cos 36.57º

Tension = 0.784 N N ÷ cos 36.57º

Tension = 0.976 N

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