How Is Tension Calculated in a Helicopter Rescue Operation?

[D. Tran]

1. "A 2800 kg helicopter with a cable and harness is rescuing an injured snowboarder. The harness and snowboarder weigh 1730 N."

The question is asking for the "tension in the cable if the snowboarder and helicopter accelerate upward at 1.5 m/s²."

2. ∑F_y=ma_y=T-


Is the equation (above) correct? What am I missing?

3. I already drew a free-body diagram at constant acceleration. With the modified equation (fitted to the problem), I plugged in the following variables:

∑F_y=T-W_S=m_Sa_y
W_S=1730 N
m_S=176.5 kg
a_y=1.5 m/s²
∑F_y=T=(176.5)(1.5)+1730N=1994.8N (final answer)

I do not know if I am doing the problem correctly. I need guidance with the problem.

Thank you for your time and help.

-D. Tran

The Attempt at a Solution

 

[PhanthomJay]

1. "A 2800 kg helicopter with a cable and harness is rescuing an injured snowboarder. The harness and snowboarder weigh 1730 N."

The question is asking for the "tension in the cable if the snowboarder and helicopter accelerate upward at 1.5 m/s²."

2. ∑F_y=ma_y=T-


Is the equation (above) correct? What am I missing?

3. I already drew a free-body diagram at constant acceleration. With the modified equation (fitted to the problem), I plugged in the following variables:

∑F_y=T-W_S=m_Sa_y
W_S=1730 N
m_S=176.5 kg
a_y=1.5 m/s²
∑F_y=T=(176.5)(1.5)+1730N=1994.8N (final answer)

I do not know if I am doing the problem correctly. I need guidance with the problem.

Thank you for your time and help.

-D. Tran

The Attempt at a Solution

Looks real good, nice work, and you avoided the temptation to factor in the helicopter's mass. Welcome to PF!

 

[kerimek]

I would like to commend you for attempting to solve this problem and providing a detailed explanation of your steps. Your approach seems to be correct, and your final answer of 1994.8N for the tension in the cable appears to be reasonable.

One thing that you may want to consider is the sign convention for the forces. Since the snowboarder and the helicopter are accelerating upwards, the tension in the cable and the weight of the snowboarder should both be positive forces. This means that your equation should be written as:

∑Fy = T + WS = mS * ay

Plugging in the values, we get:
∑Fy = T + 1730N = (176.5kg) * (1.5m/s^2)

Solving for T, we get:
T = (176.5kg) * (1.5m/s^2) - 1730N = 1994.8N

This is the same answer that you obtained, but with a slightly different sign convention. In physics, it is important to consistently use positive and negative signs to indicate the direction of the forces.

Overall, your approach and calculations seem to be correct. Keep up the good work!