need help with this one
any help would be appreciated
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Physics help PT2
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To make calculations easier, treat the system as both of the blocks. Now imagine getting rid of the pulley in the middle as the tension does not matter (it actually cancels out). You would have two blocks attached by a string in the middle each with a force acting left and right. m2 will be the block on the right (positive direction), m1 the block on the left (negative direction). There are a total of two forces acting on the object, either kinetic/static friction on m1 ,depending on whether the block moves, and the force of gravity on m2.
The diagram should look something like F(friction) <-[m1]------[m2]-> F(gravity on m2). Now the two parts of the problems tests your knowledge on the two types of friction. You should know that on a surface with friction, a stationary object will experience static friction in the opposite direction when a force is acted upon it. The object will only accelerate from rest when that force is greater than the threshold of static friction. F > μ(s)*F(N) for the object to accelerate (imbalanced forces). If the force acting upon the object is less than or equal to the force of static friction, the object will not accelerate from rest (Note: normally, when two forces are equal and in the opposite direction, an object can either be nonmoving or be moving at a constant velocity with no acceleration. This is not the case for static friction, which can only be overcome when the force is greater than friction). On the other hand, kinetic friction will occur when an object is already moving.
Now with that in mind, the force of static friction is acting on the object in the first part of the problem when the system is at rest. μ(s)F(N) = m1 g μ(s) = 54.88 N. On the right side, the force of gravity only acts upon m2 (mass of string/pulley negligible). F(g m2) = m2 g = 44.1 N. Since F(static friction on m1) > F(g m2), the entire system will remain at rest with no acceleration (The system does not accelerate to the left even if the force on the left is greater than the right because friction is not a force that accelerates an object FROM rest, rather it accelerates an object TO rest.)
The second part of the problem now deals with kinetic friction, which again only acts upon m1 after m2 is stated to be accelerated downwards by some external force. Now use Newton's 2nd Law again. Fnet = ma. You try figuring out the forces this time for the left hand side (hint: there are two forces). It is also important to note here that on the right hand side, 'm' is the mass of the entire system (both blocks, string mass negligible) because the entire system is going to accelerate from rest, not just one of the blocks. Add the masses of the blocks together and solve for acceleration when done with the force calculations.
Note how the coefficient of static friction is > the coefficient of kinetic friction. This is usually the case, as it is harder to start moving a heavy object than continue to move a heavy object. Once crossing the static friction threshold, moving it requires less force.
Now with that in mind, the force of static friction is acting on the object in the first part of the problem when the system is at rest. μ(s)F(N) = m1 g μ(s) = 54.88 N. On the right side, the force of gravity only acts upon m2 (mass of string/pulley negligible). F(g m2) = m2 g = 44.1 N. Since F(static friction on m1) > F(g m2), the entire system will remain at rest with no acceleration (The system does not accelerate to the left even if the force on the left is greater than the right because friction is not a force that accelerates an object FROM rest, rather it accelerates an object TO rest.)
So even with the static friction coefficient, there is no acceleration? This would make the answer 0 for the first part then, right?
The second part of the problem now deals with kinetic friction, which again only acts upon m1 after m2 is stated to be accelerated downwards by some external force. Now use Newton's 2nd Law again. Fnet = ma. You try figuring out the forces this time for the left hand side (hint: there are two forces). It is also important to note here that on the right hand side, 'm' is the mass of the entire system (both blocks, string mass negligible) because the entire system is going to accelerate from rest, not just one of the blocks. Add the masses of the blocks together and solve for acceleration when done with the force calculations.
Could you solve this one out? I am still not understanding
- Yes. Static friction prevents gravity from accelerating the blocks because the force of static friction is greater than that of gravity.
The coefficient is kinda like a constant, sitting in between F(fs) = μ(s) * F(N), and showing that static friction and normal force are proportional: the higher the normal force and thereby weight, the more friction (thats why a heavy object is harder to push along a surface, requiring more force to overcome static friction). I would say that just keep in mind the higher the coefficient (usually a number between 0 and 1), the higher the force of friction. But, for the most case, it is primarily used for calculation purposes and nothing more, so don't worry too much about it.
- Use the diagram. Treat the force of gravity as positive and kinetic friction as negative. Fnet = ma. F(gravity on m2) - F(kinetic friction on m1) = (m1+m2) * a. plug the numbers in from there. You should know the formulas for the forces, but the ap test will provide you with them so you dont have to necessarily memorize them 100%.
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