A 15000-kg helicopter is lowering a 3800 -kg truck by a cable of fixed length. T
ID: 2142487 • Letter: A
Question
A 15000-kg helicopter is lowering a 3800-kg truck by a cable of fixed length. The truck, helicopter, and cable are descending at 12.0 m/s and must be slowed to 5.00 m/s in the next 44.0 m of descent to prevent damaging the truck. Assume a constant rate of slowing.
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A 15000-kg helicopter is lowering a 3800-kg truck by a cable of fixed length. The truck, helicopter, and cable are descending at 12.0 m/s and must be slowed to 5.00 m/s in the next 44.0 m of descent to prevent damaging the truck. Assume a constant rate of slowing. Draw the free body diagram of the truck. Determine the tension in the cable. Determine the lift force of the helicopter blades.Explanation / Answer
a)
From kinematic equations we have:
Vf2 = Vi2 + 2a?y
Vf2 - Vi2 = 2a?y
Since we are solving for acceleration divide off 2?y
(Vf2 - Vi2) / (2?y) = a
a = [(5m/s)2 -(12m/s)2] / (2*44 m)
a = -1.352 m/s2 (note acceleration is negative, means downward as we expected since the load is being lowered)
Plug acceleration back into your tension equation from step 3
+ T = mT(-a) (mass of truck)
T = 3800 kg * [-(-1.352 m/s2)] = +5138.636 Newtons
T = 5158.64 N =5.159 kN
b) F lift = FL = ?
From step 1, we have ?Fy = + 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/s2.
So what is the difference between those 2 values? It's +9.8 + (-1.3522) = +8.4478m/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 = (+8.4478m/s2) (15 000 kg + 3800 kg) = 158818.64 N.
FL = 158818.64 N = 158.819 kN
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