MCAT Physics Chapter 2 Practice Passage

ryannellie
Posts: 2
Joined: Mon Aug 12, 2019 8:06 pm
Location: Oklahoma

MCAT Physics Chapter 2 Practice Passage

Hello! I have a question about the answer to the practice passage for Chapter 2 of the Physics textbook. It says that force of gravity acting on the beam is 2kg (-9.8) = ~ -20N. Where are is the 2kg coming from? The second mass is listed at 1kg, the string is massless and we are trying to determine the mass of the first block.

Thank you for any help!
NS_Tutor_Mathias
Posts: 616
Joined: Sat Mar 30, 2019 8:39 pm

Re: MCAT Physics Chapter 2 Practice Passage

I'm not sure where you're getting 1kg from:
Students in a biophysics class used a hinged beam setup (Figure 1) in order to simulate forces present in the human
arm. A horizontal beam (L = 0.5 m) with a mass (Mb) of 2.0 kg, was attached to a frictionless hinge. Te hinge was
also attached to a vertical beam which has a frictionless wheel at its top.
Noor0110
Posts: 6
Joined: Wed Mar 11, 2020 3:36 pm

Re: MCAT Physics Chapter 2 Practice Passage

Hi there, I'm struggling with question 6 from the passage. It says "If the string is cut, releasing M1, what will happen to the horizontal beam?" I understand that the beam will rotate because the string is cut and there is no CCW Torque to balance the CW Torque. However, I do not understand why the center of mass translates down and then left?
NS_Tutor_Mathias
Posts: 616
Joined: Sat Mar 30, 2019 8:39 pm

Re: MCAT Physics Chapter 2 Practice Passage

Consider the initial center mass:
On the horizontal and vertical axes, that is the center of mass of the beam added to the center of mass of the two weights. If you cut the string, the center of mass of the system is now one weight and the center of mass of the beam. So it is just asking you what the center of mass of the new system (without the right-hand weight) would be.
mfarnell
Posts: 1
Joined: Thu Apr 30, 2020 2:11 pm

Re: MCAT Physics Chapter 2 Practice Passage

I am also confused about the explanation for question 2:
What is the approximate tension in the string when the system is at equilibrium in experiment 1?

The explanation mentions a lot of different forces but is not clear in what it is referring to. Could you try explaining another way? Thanks.
NS_Tutor_Mathias
Posts: 616
Joined: Sat Mar 30, 2019 8:39 pm

Re: MCAT Physics Chapter 2 Practice Passage

The counterclockwise torque has to be equal in magnitude to the clockwise torque. This is a condition necessary for rotational equilibrium.

Rather than solving and arriving that exactly the conclusion this problem arrives at, let me outline the steps to solving this problem and then discuss a small oversight in this old book problem:

1. Set both torques equal
2. Realize that the source of the CCW-torque must be the vertical component of the tension in the string, exerted at the attachment point to the beam
3. Find out how big that vertical component must be
4. Use trigonometry to solve for tension (Fvert/hypothenuse = cos(30) -> hypothenuse = Fvert/cos(30) )

Now, there is a small hiccup in this problem, in that the mass attached at the end of the horizontal beam is accounted for - even though the mass of this object is simply not given. This definitely contributes to the confusion for the solution pathway, because only M1 is given and only in experiment 2.

So here is a suggestion: Give solving this another shot, and take it from me that M2 = 1 kg. Let me know how it goes, and if after a bit of work on your own you still feel more clarification could be helpful, I'll be happy to provide a sketch