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Physics 18 Online
OpenStudy (anonymous):

A helicopter flies at a constant altitude towing an airborne 100 kg crate as shown in the diagram. The helicopter and the crate only move in the horizontal direction and have an acceleration of 3.0 m/s2. Find the magnitude of the tension in the cable (in Newtons). Ignore the effects of air resistance.

OpenStudy (matt101):

You didn't provide a diagram, but I'm assuming the cable is at an angle pointing away from the direction of motion. That means you can deconstruct the tension into horizontal and vertical components. The vertical component of the tension is the force of gravity on the crate, while the horizontal component of the tension is the forward force applied by the helicopter to produce the given acceleration: |dw:1444363914868:dw| You can now set up an equation based on the Pythagorean Theorem to solve for the tension: \[T^2=(mg)^2+(ma)^2\]\[T=\sqrt{m^2g^2+m^2a^2}\]\[T=m \sqrt{g^2+a^2}\]\[T=100 \sqrt{9.8^2+3^2}\]\[T=1025\] So the tension in the cable is about 1025 N! Does that make sense?

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