I was working on a lab in class drawing Free Body Diagrams. The problem required we drew an FBD of a ball that is in the motion of being thrown. I drew a slightly diagonal line labelled Applied Force, a vertical line straight down labelled Gravity, and a line opposite of Gravity labelled Normal Force. I was told that there is no normal force at all in this situation, and that the only two forces are gravity and applied force. I was confused because I thought that the ball was being held up by the hand (acting as a surface) which provided a normal force while still being pushed by the hand in the positive direction resulting in an applied force. So in this situation, would the ball have any normal force at all acting on it?
[Physics] Does a ball have a normal force exerted on it if the ball is thrown
forcesfree-body-diagram
Related Solutions
Firstly, the 'normal' force is a contact force which opposes two surfaces being forced together. It is not caused by gravity, but it can be the response to gravity (eg when gravity pushes a book into contact with a table). It is caused by inter-molecular repulsive forces.
Secondly, the van der Waals force is an attractive force. This is the only upward force here, pulling the gecko towards the ceiling.
However, does this mean that the pads on the gecko's feet are so finely controlled that the van der Waals force exactly balances the weight of the gecko? ie If the gecko has a larger breakfast than usual the pads somehow exert a stronger van der Waals force to compensate? I am not sure this is the case.
Even if the gecko can control the van der Waals force, I doubt that it can be done so precisely. I think it more likely that the van der Waals attractive force is always slightly greater than gravity, pulling the gecko into the ceiling. The usual repulsive inter-molecular forces then come into play to oppose the net upward force on the gecko. The normal contact force is reactive and adjusts to provide a perfect balance of forces.
So in addition to its weight (downward) and the van der Waals force of attraction from the ceiling (upward) I think there is also a small additional 'normal' (in both meanings of the word!) contact force from the ceiling acting downward on the gecko. This is where I disagree with Harm Moolenaar. (I do not understand what Joce is saying - possibly the same as Harm?)
To some extent the issue might be one of semantics - ie whether 'van der Waals force' includes the inter-molecular repulsion or is separate from it. Nature does not distinguish one force from another, only we do that, for our own convenience. So I think it is an issue of how forces are distinguished and classified.
It's alright, tension is pretty subtle! Let me answer your questions a bit out of order.
- If you pick any and all points on the rope, would there be two opposing tensions at every one of those points?
- Is tension uniform throughout the rope?
You can think about the rope as a lot of tiny masses connected together by springs; this is a cheap approximation for how tension works on the atomic level, where the springs are stretching chemical bonds and the masses are atoms.
In our simple model of tension, every atom is pulled by a spring on the left and a spring on the right, with forces $T_1$ and $T_2$. Then by Newton's second law, $$T_1 - T_2 = ma$$ where $m$ is the mass of the atom. Since $m$ is very very tiny compared to the other masses in the problem, we must have $$T_1 \approx T_2.$$ Applying this to every mass, we conclude that each little spring / chemical bond has approximately the same tension, so we can simply talk about "the tension in the rope". This is a good approximation as long as the total mass of the rope is much smaller than the masses of the blocks.
How might differences in mass between object A and object B (which, sorry if the diagram was misleading in the sizes, can have any mass) play into the tension?
Well, now that we've established that there's a uniform tension $T$, we need to figure out what that tension is. The constraint here is that the rope is taut, which means that it can't be scrunching up or stretching out; that translates to the constraint that the accelerations of the two blocks are equal in magnitude. This equation determines the tension.
How does the relationship between the force of gravity on mass B and the tension in the rope play into this? Isn't the tension caused by that force of gravity? Doesn't that mean that if tensions cancel, the force of gravity's effect is canceled as well? (I have also been told that the blocks will have the same magnitude of acceleration. Why is this?)
No, the tension isn't equal to the weight of block B, it's whatever is necessary to satisfy the above constraint. For example, suppose that block B was very very heavy, so the acceleration of the whole system will be close to $g$. In this case, the tension is actually quite small compared to the weight of block B, because you only need a little tension to make block A have the same acceleration.
(In fact, as block B gets infinitely heavy, you can show that the tension doesn't go to infinity -- instead, it becomes the weight of block A! It's neat to try to prove this, and see how it works.)
Does the pulley affect tensions? For example, we know that there's a positive tension affecting mass A. Is there still a positive tension in existence on the other side of the pulley, or just the negative tension that's acting on B? Might there be some sort of effect whereby two sets of opposite tensions, one set on each side of the pulley, cancel each other out?
This is a little tricky to word. The tensions of the two tiny springs attached to each atom approximately cancel out, as shown above. But that doesn't mean that the tension is zero -- all of those springs are still stretched.
Since the pulley is frictionless, it doesn't have any effect except that it 'turns around' the tension. You can show this by considering the three forces on each atom (two springs, one normal force) which gives $T_1 \approx T_2$ as before.
Anyway, I've been told that the net force is the force of gravity on mass B (the hanging one). But I don't really understand why, even after extensive discussion with various people.
That's not your fault, the question is just worded badly. There are lots of forces involved in this problem acting on different things: gravity on both blocks, normal on one block, and normal from the pulley on the rope. It's not very clear what "the" net force even means.
Best Answer
Let's review what a normal force is.
Solid objects are characterized by an inability to occupy the same volume of space as other matter. The microscopic reasons for this behavior are very complicated, but the macroscopic result are simple to describe.
A normal force is the force that enforces the "no occupying the same volume" rule for solids and as such it always acts perpendicularly to the surface of contact and points pushes the two objects away from each other (this is a Newton's 3rd Law pair).
Somethings that are not characteristics of normal forces
Now, on to your situation.