i need help with part c i believe i answered part a and b correctly we can model a pine tree in

First, we know that the maximum torque the tree can withstand is 105 N-m. This torque is created

I need help with part c i believe i answered part a and b...

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.) a. If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3. Hint: Use the equation for the drag force. Drag force equation: FD = 1/2pv^2CdA FD = 1/2(1.2 kg/m^3)(6.5 m/s)^2(0.50)(9 m^2) FD = 114.075 N p = density v = velocity Cd = drag coefficient A = area b. What torque does this force exert on the tree, measured about the point where the trunk meets the ground? T = Fxr = Frsinθ T = (114.075 N)(7 m)sin(90) T = 798.525 N-m c. Suppose that the maximum torque that this pine tree can withstand without toppling is 105 N-m, what wind speed will cause the tree to fall over?

Transcript

00:04 In this problem, we're going to deal with drag force and torque.

00:09 It's talking about a pine tree.

00:12 So if you look over here on the right hand side, i have, of course, a pine tree, but you'll have to forgive it because i don't know what a pine tree with a condensed canopy looks like.

00:23 So the canopy right here was given that it was nine meters squared for the area.

00:30 The wind is going to blow on center to the canopy at six point.

00:39 0 .5 meters per second, and it's 7 meters from the center of the canopy to the ground.

00:50 So i have most of this information written over here on the left.

00:57 This is the equation for drag force.

01:01 It's one half, the density of air, which is 1 .2 kilogram per meter cubed, time velocity squared, and the velocity was given to me in the problem.

01:12 As 6 .5 meters per second.

01:18 C sub d is the coefficient of drag, and that was given as 0 .5.

01:25 And the a is area, which is the area that the velocity is felt, which is the canopy at 9 meters squared.

01:37 So the beginning of this problem is a matter of putting all these formulas or all these numbers into my formula.

01:46 So i have the drag force will be equal to one -half, density of 1 .2, velocity squared, which is 6 .5 squared, coefficient of drag, which is 0 .5, and area of 9.

02:08 I plug it into the calculator all in one step and you get 11 .075 newtons.

02:21 So that's the first part of our question.

02:24 What was the drag force? the second part is what is the torque that is felt by this pine tree and the torque is the rotational force.

02:36 So my wind is blowing in a horizontal direction, but because the tree is fixed into the ground, it has a rotational force.

02:46 So the equation for torque is that drag force times the distance that it's felt.

02:56 And my distance is going to be just the seven meters.

03:01 Because there is no x component, it's just all in the y.

03:05 If you wrote this out using sign, we would have force of drag times the distance, times sign of 90.

03:20 And again, that's because it has felt over here on the right by my tree, it's felt at a 90 degree angle.

03:29 So the wind is coming in at the x and it's felt at a fixed point on the y axis.

03:37 And sign of 90 is equal to 1, which is why we can cancel it off and just use the drag force of 114 .075 times the distance of 7.

03:55 And we get a torque equal to 798 .53...

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