Here's the question you clicked on:
brinethery
A man with mass m1 = 53.0 kg stands at the left end of a uniform boat with mass m2 = 172.0 kg and a length L = 2.5 m. Let the origin of our coordinate system be the man’s original location as shown in the drawing. Assume there is no friction or drag between the boat and water. After the man walks to the right edge of the boat, what is the new location the center of the boat? Now the man walks to the very center of the boat. At what location does the man end up?
no displacement of the man(position)?
Sorry for the misunderstanding. He's initially at the left side, and then he walks to the right side of the boat.
The center of mass of the man+boat system must not move because internal forces cannot accelerate a system. let m1=mass of the man m2=mass of the boat Let the x-position center of mass of the man be x1 (initially 0) x2= center of mass of the boat (= +L/2 initially) the x-co-ordinate of the center of mass of the boat+man system is given by \[\Large x=\frac{m_1x_1 + m_2x_2}{m_1 + m_2}\] Once you find this for the initial condition, it doesn't change for the rest of the problem. Let's concentrate on the first question. After the man moves to the other side of the boat, the distance between him and the center of the boat is exactly L/2. i.e. \( X_1-X_2=L/2=1.25 m.\) use the equation of the COM of the system to get another equation in \(X_1\) and \(X_2\). i.e. \(\Large 1.91=\frac{53(X_1)+172(X_2)}{53+172}\) i.e \(\large 430=53(X_1)+172(X_2)\) Solve those two equations to get \(X_2\), the answer to the first question.
to answer the second question, repeat the same steps: 1. Get one equation by applying the COM formula. 2. The second equation is easy: since the man is standing on top of the center of the boat, his COM and the boat's COM coincide. So, \(X_1-X_2=0\; or\; X_1=X_2\).
For the first question, I did not come out with the right answer. x1 = 2.87 x2 = 1.62 Could someone please help me further? I do not understand any of this.
I think that's correct. I got the same thing.
Does your book say otherwise?
I put both answers in to smartphysics and it's saying it's wrong.
I am discouraged :-(
Hey I have to leave for class but I will be back later.
The answer is absolutely right. The problem may lie with you entering the answer. Did they ask you to enter the answer with a certain number of decimals (or significant digits) ?
That's the problem with smartphysics. Most of the time, they don't say anything. Just a little red x if it's wrong, hehe. I have to use this website for the rest of the year. Dammit :-(