Given:

Assume up is +, down is -

Mass helicopter, m_{H }= 15 000 kg

Mass truck, m_{T} = 4300 kg

Velocity initial, V_{i} = 17m/s

Velocity final, V_{F} = 5m/s

Change in vertical distance, Δy = 56m

A constant rate of slowing = constant acceleration = hint to use kinematic equations.

~~~~~~~~~~~~~~~~~~~~~~~~

a) Tension = T = ?

b) F lift = F_{L} = ?

a)

Step 1.

Using Newton's 2nd law, ΣF =ma, we sum the forces in the vertical direction. It would help if you drew a picture. In the picture, you would have FL is up, T is up.

FgH (force of gravity of the helicopter, or weight of the helicopter) is down since gravity is pulling down,

FgT (force of gravity of the truck, or weight of the truck) is down since gravity is pulling down, and

Fd (force of descend or how the helicopter is moving the load to the ground) is down.

So, mathematically it looks like this:

ΣFy = + FL - Fd -FgH - FgT + T = m (-a) since we are moving down

Step 2.

From the above, the first term, lift, cancels the next 3 terms - descend & both weights. So the tension in the cable only depends on the load. You can also think of it visually. The cable box is installed inside the helicopter. It does not support the helicopter weight and is independent of anything else besides the load since no one is touching the cable. You can also draw a free body diagram for just the load and still come up with the same equation:

+ T = mT(-a) (mass of truck)(negative acc. since the movement is down)

Step 3.

From kinematic equations we have:

Vf^{2 }= Vi^{2} + 2aΔy

Vf^{2} - Vi^{2} = 2aΔy

Since we are solving for acceleration divide off 2Δy

(Vf^{2} - Vi^{2}) / (2Δy) = a

a = [(5m/s)^{2} -(17m/s)^{2}] / (2*56 m)

a = (-264 m^{2}/s^{2}) / 112 m

a = -2.357 m/s^{2 }(note acceleration is negative, means downward as we expected since the load is being lowered)

Step 4.

Plug acceleration back into your tension equation from step 3

+ T = mT(-a) (mass of truck)

T = 4 300 kg * [-(-2.357 m/s^{2})] = +10 135 Newtons

**T = 10 135 N**

** **

b) F lift = FL = ?

From step 1, we have ΣFy = + FL - Fd -FgH - FgT + T = m (-a) since acc. is down

remember, force of lift is the force exerted upward to counteract the weight of the helicopter, truck, and descend.

Mathematically

+ FL = + FgH + FgT + Fd (we transferred terms to the other side of equal sign, so they became positive)

more specifically we write

+FL = mH (+g) + mT(+g) + (mH+mT)(+a)

Here, the first term on the right is weight of helicopter. gravity is positive because we moved terms to the other side of equal sign.

2nd term on the right is weight of the truck.

3rd term is force of descend.

So, simplifying our equation factoring out g in first 2 terms

FL = g(mH + mT) + (mH+mT)(a)

From algebra, simplifying by combining common terms,

FL = (g+a)(mH+mT)

Note the first term. It is acceleration during descend. Gravity is always pulling down with a value

-9.8m/s2, but we transferred it to the right of the equal sign, so it is positive +9.8m/s^{2}.

So what is the difference between those 2 values? It's +9.8 + (-2.357) = +7.443m/s2. You can interpret it as if the force of gravity pulls with a lesser value now since the force of lift opposes gravity.

So

FL = (+7.443m/s^{2}) (15 000 kg + 4 300 kg) = 143 650 N.

**FL = 143650 N**

Hope that helps.

## Comments

Oh yes, I agree. I forgot the weight of the truck for the tension.