Forces involved in strapping something down 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?
 A: 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.
A: If you strap the cylinder to the bed exactly in the center, the inward forces will cancel each other out due to the symmetry of the situation. (remember that any 'diagonal' forces can be decomposed into a 'vertical' component and a 'horizontal' component) If you don't strap it to the bed dead in the center, there will be horizontal components of force on the cylinder, attempting to shift it to the center.
If there's no friction between the cylinder and the bed, these forces will succeed to push it to that center. However, without any friction, the cylinder will overshoot and in fact it will keep overshooting again and again, effectively oscillating about the center of the bed.
If there is some friction, but not too much, the horizontal forces pushing the cylinder in towards the center of the bed will do so. There will be some overshooting as well, for a while. But eventually the friction will win and the cylinder will come to a halt near the center.
If there is too much friction, the horizontal components of force due to the rope tension will not cancel each other out, but the residual force will be cancelled out by the (static) friction and the cylinder will not be moved.
In any realistic case, there is always friction. So in the end, while there are always horizontal components of force on your cylinder, the resulting horizontal force component will vanish in the symmetric case. In the non-symmetric case, the resulting horizontal force component will be non-zero and attempt to get the system into the symmetric case. How well it succeeds in doing so depends on how much friction there is.
