In a rescue, the 74.0 kg police officer is suspended by two cables, as shown in

ID: 1354210 • Letter: I

Question

In a rescue, the 74.0 kg police officer is suspended by two cables, as shown in the figure below.(Figure 1)

Part A

Find the tension in the left cable.

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Part B

Find the tension in the right cable.

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Figure 1 of 1

In a rescue, the 74.0 kg police officer is suspended by two cables, as shown in the figure below.(Figure 1)

Part A

Find the tension in the left cable.

Tleft = N

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Part B

Find the tension in the right cable.

Tright = N

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Figure 1 of 1

Let tension in left cable be T1 and right cable be T2

Vertical components of the tension are T1 sin 35 and T2 sin 48 and these 2 balance the weight

=> T1 sin 35 + T2 sin 48 = 74g = 725.2 N (eq 1)

Horizontal components are equal and opposite

=> T1 cos 35 = T2 cos 48

=> T1 = 0.817 T2 (eq 2)

Replacing in eq 1

0.817 T2 sin 35 + T2 sin 48 = 725.2

=> 1.21 T2 = 725.2

=> T2 = 598.5 N

We know from eq2, T1 = 0.817 T2 = 0.817*598.5 = 488.98 N

So tension in left cable is 488.95 N and right cable is 598.47 N

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