Calculating the Center of Mass for a Dog on a Flatboat

AI Thread Summary
The discussion revolves around calculating the new position of a dog on a flatboat after it walks toward the shore. Initially, the dog is 25.5 m from the shore and moves 6.0 m toward it. The key point is that the center of mass of the dog-boat system remains unchanged due to the lack of friction. As the dog moves closer to the shore, the boat moves away, and the distance the boat moves must compensate for the dog's movement. The final position of the dog relative to the shore can be determined by understanding this balance in the center of mass.
Punchlinegirl
Messages
221
Reaction score
0
A dog, with a mass of 12.0 kg, is standing on a flatboat so that he is 25.5 m from the shore. He walks 6.0 m on the boat toward the shore and then stops. The boat has a mass of 46.0 kg. Assuming there is no friction between the boat and the water, how far is the dog from the shore now?
i used x_cm = m_1 x_1 + m_2 x_2 / m_1 + m_2
x= (12)(25.5) + (46)(6) / 12+46
and got 10.0 m
Which wasn't right.. can someone please help me?
 
Physics news on Phys.org
The key is that the center of mass of the "boat + dog" system cannot change. So, if the dog ends up X meters closer to shore, how much further from the shore must the boat end up (in terms of X)? Now make use of the fact that you know how far the dog moved with respect to the boat to solve for X. (To help visualize what's going on, pretend the dog was standing on a line painted on the boat. The dog moves toward the shore; the boat--and line--moves away from the shore. But you know how far apart the dog moved from the line.)
 
Ok I get it now. Thank you!
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Is this center of mass question missing information?
Replies
26
Views
5K
How Does the Dog's Movement Affect Its Distance from the Shore?
Replies
2
Views
5K
Newtons Second Law For A System Of Particles
Replies
1
Views
3K
Calculating Solar Mass using peak Doppler shifts
Replies
7
Views
2K
Center of Mass boat distance from shore what am I doing wrong?
Replies
1
Views
3K
Finding the final distance of the dog from the shore
Replies
7
Views
8K
Center of Mass - Man on Boat Homework help
Replies
7
Views
7K
Center of Mass of open top cylinder
Replies
9
Views
5K
Help with Problem: Dog on Flatboat - Distance from Shore
Replies
2
Views
3K
How Far From Shore Is the Dog After Walking on the Boat?
Replies
3
Views
7K

Hot Threads

Recent Insights

Back
Top