Say I have a hollow cylinder and I wanted to strap it down to the bed of a truck. I would tension the strap on one end, and it would exert a force on the cylinder. My intuition tells me that the strap would crush the hollow cylinder down toward the truck bed, but when I think about it, there are inward forces perpendicular to the truck bed caused by the straps on the cylinder as well. Is this correct thinking, or are all of the forces only vertical?
[Physics] Forces involved in strapping something down
forces
Related Solutions
Yes, the force points along the vector of the relative velocity between the object and the air.
Quadratic drag is an interesting phenomenon. You have to calculate the net velocity vector (which includes a horizontal and vertical component) and compute the force along that axis; when you then decompose it into horizontal and vertical components you will find that the vertical drag is greater because of the cross wind. This is not an intuitive result!
Not easy to explain intuitively, but I'll give it a go. Let's use the following 3d model of the system at hand for reference (the Mathematica code used to make this model can be found in this gist):
Let in the above $\hat z$ be the direction from the base of the structure to the red sphere, and $\hat x$ the unit vector going from the red sphere towards the gyroscope.
The red arrows represent the velocity of various parts of the rotating gyroscope. The green arrows represent the effective action of gravity on the center of mass of the rod, and the corresponding counterforce provided the hinge.
The combined action of the two green arrows generates a torque on the rod + gyroscope system, trying to push it down. But the only way for this to happen, it for the gyroscope to rotate in the $\hat x \hat z$ plane:
The cyan arrows show the corresponding forces that this rotation induces on the upper and lower points of the gyroscope. Now, remember that the cyan arrows represent forces, while the red ones velocities. The cyan arrows will induce an acceleration on the the various points of the gyroscope. In particular, they will induce the red arrows to change direction. In the model this is shown for the upper and lower points, with the orange arrows represnting the modified velocity vectors in those points.
As you can see from there, the new velocity vectors correspond to the ones you have when the gyroscope changes its direction following the precession motion. Basically, the action of the cyan forces on the red velocities is what causes the precession motion.
Now for the nutation, we have to note that the cyan arrows above are not actually quite correctly drawn. The motion of the gyroscope falling dawn more correctly corresponds to the cyan arrows being slightly tilted towards the ground. This is of course to be expected: after all, if the gyroscope wasn't rotating, it would just fall down. This is why even though there is precession, the gyroscope does go down a bit.
This acceleration towards the ground, because of the same mechanism explained above, also induces a faster precession. But an accelerated precession now causes a counter-falling reaction, for reasons similar to the ones causing the precession itself.
An accelerating precession motion means that there are effectively force vectors pushing the gyroscope to rotate in the $\hat x \hat y$ plane. Again, this corresponds to force pushing different points of the gyroscope in different directions. In the following, the cyan arrow show the forces acting on two points:
Again, these cyan forces will induce a change in the direction of the corresponding red velocities, and the orange arrows show what the velocity will be soon after. As you can see, the new velocities are those corresponding to the gyroscope direction going upwards, thus generating the "counter-falling" phenomenon that is in this case the nutation.
Best Answer
If you neglect friction, the strap presses on the cylinder, and exerts a force perpendicular to the contact surface. So the infinitesimal length of strap sitting on the highest point of the cylinder exerts force downward, but at any other point, there will be a horizontal component as well. If the configuration of your strap is symmetrical, any horizontal force exerted on one side of the cylinder will be compensated by an equal, opposite force exerted on the symmetrical side.
So the overall force exerted on the cylinder has no horizontal component, only vertical, and equal to $2T \sin \alpha$, where $T$ is the tension of the strap, and $\alpha$ the angle the ends of the strap make with the horizontal. This is also the reaction that the truck bed will exert on the cylinder from below.