A man ties one end of a strong rope 8.00 m long to the bumperof his truck, 0.500

Question

A man ties one end of a strong rope 8.00 m long to the bumperof his truck, 0.500 m from the ground, and the other and to avertical tree trunk at a height of 3.00 m. He uses the truck tocreate a tension of 800 N in the rope. Compute the magnitudeof the torque on the tree due to the tension in the rope, with thebase of the tree acting as the reference point.
Known: Unknown: Torque = rFsin = ? F = 800N r = 8.00 m
I can't seem to find the correct , what I'm doingis taking the inverse cosine (2.5m/8.0m) which gives me an angle of71.8 degrees, so I am using sin(18.2) in the formula, but I get1999 Nm as my final answer which doesn't the book answer of 2280Nm. Any idea where I get wrong?

A man ties one end of a strong rope 8.00 m long to the bumperof his truck, 0.500 m from the ground, and the other and to avertical tree trunk at a height of 3.00 m. He uses the truck tocreate a tension of 800 N in the rope. Compute the magnitudeof the torque on the tree due to the tension in the rope, with thebase of the tree acting as the reference point.
Known: Unknown: Torque = rFsin = ? F = 800N r = 8.00 m
I can't seem to find the correct , what I'm doingis taking the inverse cosine (2.5m/8.0m) which gives me an angle of71.8 degrees, so I am using sin(18.2) in the formula, but I get1999 Nm as my final answer which doesn't the book answer of 2280Nm. Any idea where I get wrong?

Explanation / Answer

is the angle that the rope makes with the tree (which isvertical). By taking the inverse cosine of (2.5m/8.0m) youcan obtain this angle. So, you should be using71.8o as your . The other mistake is thatthe torque is acting about the base of the tree, so the r should be3.00m. (The distance from the bottom of the tree to the pointat which the rope is acting) By using F = 800N, r = 3.00m,and = 71.8o , you get 2280 Nm.

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