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brinethery
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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?
 one year ago
 one year ago
brinethery Group Title
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?
 one year ago
 one year ago

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vannayen Group TitleBest ResponseYou've already chosen the best response.0
no displacement of the man(position)?
 one year ago

brinethery Group TitleBest ResponseYou've already chosen the best response.1
Sorry for the misunderstanding. He's initially at the left side, and then he walks to the right side of the boat.
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
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 xposition center of mass of the man be x1 (initially 0) x2= center of mass of the boat (= +L/2 initially) the xcoordinate 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. \( X_1X_2=L/2=1.25 m.\) 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.
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
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, \(X_1X_2=0\; or\; X_1=X_2\).
 one year ago

brinethery Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
I think that's correct. I got the same thing.
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
Does your book say otherwise?
 one year ago

brinethery Group TitleBest ResponseYou've already chosen the best response.1
I put both answers in to smartphysics and it's saying it's wrong.
 one year ago

brinethery Group TitleBest ResponseYou've already chosen the best response.1
I am discouraged :(
 one year ago

brinethery Group TitleBest ResponseYou've already chosen the best response.1
Hey I have to leave for class but I will be back later.
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
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) ?
 one year ago

brinethery Group TitleBest ResponseYou've already chosen the best response.1
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 :(
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
Tough luck.
 one year ago
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