# Tough Mechanics Problem

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#### Twitch

##### Full Member
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A fire helicopter carries a 584 kg bucket at the end of a 9.3m long cable. When the helicopter is returning from a fire at a constant speed of
58.1 m/s, the cable makes an angle of 27.3 degrees with respect to the vertical. The acceleration of gravity is 9.8 m/s^2

Find the horizontal force exerted by air resistance on the bucket. Answer in units of N.

#### Shrike

##### Lanius examinatianus
15+ Year Member
Y_Marker said:
A fire helicopter carries a 584 kg bucket at the end of a 9.3m long cable. When the helicopter is returning from a fire at a constant speed of
58.1 m/s, the cable makes an angle of 27.3 degrees with respect to the vertical. The acceleration of gravity is 9.8 m/s^2

Find the horizontal force exerted by air resistance on the bucket. Answer in units of N.
This is not an MCAT problem.

However:

Air resistance is balanced by the horizontal component of the tension on the cable. Gravity is balanced by the vertical component of the tension.

T (tension on cable) = (584 x 9.8)/cos(27.3) = F/sin(27.3)

Approximating (as we would on the MCAT): 584 x 9.8/cos(30) = F/sin(30)

F = 584 x 9.8 x 0.5/.87 = 3289N (at 30 degrees)

(In practice, we'd use 10 for gravity, too.)

#### Twitch

##### Full Member
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Thanks Shrike.

I had a similar solution (2954N) but I kept wondering about the other information that was provided on the helicopter as it relates to velocity of the chopper, length of the cable etc.. The lesson learned - don't worry about extraneous data.

Shrike said:
This is not an MCAT problem.

However:

Air resistance is balanced by the horizontal component of the tension on the cable. Gravity is balanced by the vertical component of the tension.

T (tension on cable) = (584 x 9.8)/cos(27.3) = F/sin(27.3)

Approximating (as we would on the MCAT): 584 x 9.8/cos(30) = F/sin(30)

F = 584 x 9.8 x 0.5/.87 = 3289N (at 30 degrees)

(In practice, we'd use 10 for gravity, too.)