Does anyone know how to solve this? Or why the answer is C?
View attachment 39227
For reference:
https://physics.bu.edu/~redner/211-sp06/class05/static.html
Now, understanding static friction: A simple understanding of static friction is the least amount of force needed to move an object across another surface. Another way to think about it is how much force is preventing your object from moving across a surface when another force is acting on it.
The formula is as follows: Fsmax=us*N.
us=coefficiant of static friction, N is Newtons/force of the object in question, and Fsmax=total force needed to move the object/total force acting on our object.
Looking at the problem: We can rule out A and D automatically. If an object is at rest static friction equals zero, and the coefficient will never be greater than 1 if the force acting on the object is less than the weight of the object its self.
Lets assign some variables: So because we are trying to discern how much force is needed to move the 4kg object using the 2kg mass, 2kg becomes our Fsmax.
Fsmax=2.
The weight of the object in question is 4kg, so that becomes our N.
So N=4.
So we end up with: 2=us*4
We could convert the Kg to N if we wanted, and in fact posterity says we should, but we can skip this step because the math ends up getting us the same answer in any case and is much simpler this way, so we will skip.
Algebra gets us 2/4=us. Our coeficient has to be .5 then.
How do we discern if it can be less than or more than? Well, if we take .6(more than .5) and multiply by 4, we end up with a value more than 2. Remember, static frictional force is only concerned with maximums, so we can't exceed 2 in our calculations. Another way to think about this is that we can't magically change the weight that is dangling off the table. 2Kg is the max, that is ALL the force acting on our object. Meanwhile, if we take .4(less than .5) and multiply it by 4, we end up with something less than 2.
^Really, this extra step is just annoying bullshit and the answer is .5. The coeficient of static friction between the table and the 4kg block with 2kg of weight acting on it is .5 and can't exceed .5 unless something is purposefully changed, which would fundamentally change the question in itself. As long as you understand that this formula will give you the maximum possible coefficient you won't even have to think about it.