Ok, so my Ff that I got is the minimum force required to move the blocks at all? But then I don't understand how I'm supposed to solve for anything else if I'm not given any other values?When you multiply the coefficient of static friction by the normal force you have the maximum force of static friction, i.e. the value of ##f_s## at which sliding occurs. The force of static friction is "whatever is necessary to provide the observed acceleration of the system." Here the observed acceleration is provided by force P.
- What is the observed acceleration of the two-block system?
- What is the net force on block A that provides this observed acceleration?
- How large can P become before the blocks stop moving as one and slide relative to each other?
To help wrap your head around this problem, here are some hints:
(I.) Forget the given mass of block B for a second. Just for the moment, pretend that the mass of Block B is unknown. Don't worry we can put the known value back in later. But for the moment, pretend you don't know it. But keep the same coefficients of friction (i.e., 0.4 between the masses, and a frictionless surface between the ground and block B). No math on this one. Just ask yourself what you would expect if block B's mass were an extreme value.
How would the problem change if mass B were really large, say infinitely large?
How would the problem change if mass B were really small. Like say, what if mass A was made of lead, and mass B of styrofoam?
You can use this gathered intuition later as a sanity check for equation that brings about the final answer.
(II.) When analyzing your free body diagrams (FBDs), don't forget that this is a dynamics problem, not a statics problem. The sum of all forces do not equal 0. Rather the sum of all forces, [itex]\Sigma_i Fi = a \cdot \Sigma _j m_j[/itex], assuming that both blocks have the same acceleration*. This idea applies to individual blocks within the system too. So what's the overall acceleration of the system as a function of both masses and the applied force? Now do a FBD of block A. (intermediate hint: What must be the reactive force be on block A from block B, now that you know block B's acceleration?)
*[Edit: however, the frictional force applied from block B to block A must be equal-&-opposite the sum of all forces on block B, of course, because of Newton's 3rd law, and because block B is lying on a frictionless surface.]
You are missing the point that "Block B sits on a frictionless surface." If P were, say, one millionth of ##\mu_sN## would the blocks accelerate or not?
Another way to look at it is to pretend that the blocks are glued together with super glue so there is no sliding. What would their acceleration be in this case?
Now make the glue so weak that the contact is frictionless. What happens then? Answer: Block A will accelerate with acceleration ##P/m_A## and block B will not move.
Imagine now that the glue strength is between "super" and "frictionless." What force accelerates block B? Certainly not P because P acts on block A.
| # | Наименование новости | Тональность | Информативность | Дата публикации |
|---|---|---|---|---|
| 1 | 0 | 0 | 24-10-2018 | |
| 2 | 0 | 0 | 26-10-2018 | |
| 3 | 0 | 0 | 08-09-2026 | |
| 4 | 仮 | 0 | 0 | 16-02-2026 |
| 5 | 仮 | 0 | 0 | 16-02-2026 |
| 6 | 仮 | 0 | 0 | 16-02-2026 |
| 7 | Momentum vs inertia | 0 | 10 | 19-10-2026 |
| 8 | перевод координат | 0 | 0 | 18-02-2025 |
| 9 | Точность длинного висячего хода | 0 | 0 | 10-09-2025 |