TBR Torque Help, PLEASE!

[DPTinthemaking15]

Here is my dilemma. I went through the Forces and Torque section and felt fairly confident, until I reached the Rotational Equilibrium section. I do not understand what is going on when they add a fulcrum to the problem. After watching Khan Academy, AK lectures, random Youtube videos, etc. So far nothing is sticking, and this problem is giving me the biggest headache. Intuitively I understand how to get the problem, but I would like to know the math, in case I encounter a problem with similar answer choices. My issue is this: (250N)X- (300N) (6-X)=0.... I understand everything until I reach (6-X). How do you know to subtract X? This is how I would have set the problem up, but I know my answer is incorrect: (250N)R= (300N) (6).... Also, if anyone can point me towards a good source to understand Rotational Equilibrium, it would be much appreciated.

Edit:

I have one other question as well. I finished covering Phase I and understand a lot more and decided to complete phase II (It went A LOT better). Anyways, can anyone explain this problem as well.

As a person jumps off of the floor, what is true in terms of the forces involved?

a. The reactionary normal force of the floor is less then the weight force of the person.
b. The reactionary normal force of the floor is equal to the weight of the person.
c. The reactionary normal force of the floor is equal to the force generated by the muscle groups causing plantar flexion.
d. The reactionary normal force of the floor is greater than the force generated by the muscle groups causing plantar flexion.

I chose answer choice C. And apparently D is the correct answer.

Here is TBR's explanation: In the case of a person jumping off of the floor, their center of mass experiences an acceleration upwards from rest until reaching some max acceleration after which acceleration decreases. So for a period of time at the start of the jump, the force up (in this case normal force) must exceed the sum of the forces down (weight plus whatever force is associated with the muscles pushing down on the feet). This tells us that the reactionary normal force cannot be less than or equal to the weight of the person and that it must be greater than, rather than equal to, the force generated by the muscle groups causing plantar flexion. This eliminates A,B, and C.

If acceleration and force are proportional to one another, why would the reactionary forces differ?

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[MCAT Self Prep]

In regards to your first question, rotational torque questions have the following set up: clockwise torque = counterclockwise torque. Because torque equals F (force) * r (distance from fulcrum), you will set up the equation as follows: (250)(X)= (300)(6-X). The reason we use 6-X for the right side is that we don't know the distance between the fulcrum and the 300N weight. All we do know about it is that it will be equal to 6 minus the distance between the fulcrum and the 250N weight (X). For instance, if X equals 2, then the distance between the fulcrum and the 300N weight must be equal to 6-2 = 4. I hope this helps!

 

[DPTinthemaking15]

In regards to your first question, rotational torque questions have the following set up: clockwise torque = counterclockwise torque. Because torque equals F (force) * r (distance from fulcrum), you will set up the equation as follows: (250)(X)= (300)(6-X). The reason we use 6-X for the right side is that we don't know the distance between the fulcrum and the 300N weight. All we do know about it is that it will be equal to 6 minus the distance between the fulcrum and the 250N weight (X). For instance, if X equals 2, then the distance between the fulcrum and the 300N weight must be equal to 6-2 = 4. I hope this helps!

Thank you so much! This makes sense now.