U2-2-2- QUESTION 1 A 67 kg police officer hangs from ropes as shown. Find the te

Question

U2-2-2-

QUESTION 1

A 67 kg police officer hangs from ropes as shown.

Find the tension in each rope.

Consider positive going right and upward with x-dir horizontal and y-dir vertical.

Look at the forces on the police officer only.

Angle that rope A makes with the horizontal x-axis at the police officer's head in degrees is

QUESTION 2

Angle rope B makes with the horizontal x-axis at the police officer's head in degrees is

QUESTION 3

Force due to the weight of the person (N) (Can be positive or negative)

QUESTION 4

Magnitude of the tension in rope A (N) (Positive)

QUESTION 5

X-component of the tension in rope A (N) (Can be positive or negative)

QUESTION 6

Y-component of the tension in rope A (N) (Can be positive or negative)

QUESTION 7

Magnitude of the tension in rope B (N) (positive)

QUESTION 8

X-component of the tension in rope B (N) (Can be positive or negative)

QUESTION 9

Y-component of the tension in rope B (N) (Can be positive or negative)

Explanation / Answer

Assume that rope A is on the left side of triangle.
For rope A:
Vertical = T1 * sin 35
Horizontal = -T1 * cos 35
This is negative because its direction is left.

For rope B:
Vertical = T2 * sin 48
Horizontal = T2 * cos 48
Weight = 67 * 9.8 = 656.6 N

Eq#1: T1 * sin 35 + T2 * sin 48 = 656.6
-T1 * cos 35 + T2 * cos 48 = 0
T1 * cos 35 = T2 * cos 48
Eq#2: T1 = T2 * (cos 48 ÷ cos 35)

Let’s substitute this for T1in Eq#1.

T2 * (cos 48 ÷ cos 35) * sin 35 + T2 * sin 48 = 656.6
T2 * cos 48 * tan 35 + T2 * sin 48 = 656.6
T2 = 656.6 ÷ (cos 48 * tan 35 + sin 48)
T2 is approximately 541.9 N.

T1 = [656.6 ÷ (cos 48 * tan 35 + sin 48)] * (cos 48 ÷ cos 35)
T1 is approximately 442.65 N

If these answers are not correct, the angles must be measured from vertical.
For rope A:
Vertical = T1 * cos 35
Horizontal = -T1 * sin35
This is negative because its direction is left.

For rope B:
Vertical = T2 * cos 48
Horizontal = T2 * sin 48
Weight = 67 * 9.8 = 656.6 N

T1 * cos 35 + T2 * sin 48 = 656.6

T1 * sin35 = T2 * sin 48
T1 = T2 * sin 48 ÷ sin 35
Substitute this for T1.

T2 * sin 48 ÷ sin 35 * cos 35 + T2 * sin 48 = 656.6
T2 * (sin 48 ÷ sin 35 * cos 35 + sin 48) = 656.6
T2 = 656.6 ÷ (sin 48 ÷ sin 35 * cos 35 + sin 48)
T2 is approximately 363. 9 N
T1 = [656.6 ÷ (sin 48 ÷ sin 35 * cos 35 + sin 48)] * sin 48 ÷ sin 35
T1 is approximately 471.4 N

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