Solve Third Law Problem: 833N Diver from 375kg Boat w/ 2.25m/s Velocity

  • Thread starter Thread starter oMovements
  • Start date Start date
  • Tags Tags
    Law Third law
AI Thread Summary
A diver weighing 833 N jumps from a 375 kg boat with a horizontal velocity of 2.25 m/s. The initial calculations incorrectly included gravitational force, which should only relate to the diver's mass. After correcting the approach, the net force on the boat is determined to be -382.5 N, leading to a final velocity of 0.51 m/s for the boat. The discussion emphasizes that in third law problems, the interactions between the diver and the boat are distinct from those involving gravitational forces. Understanding vector direction is crucial in these calculations.
oMovements
Messages
28
Reaction score
0
Homework Statement
A diver with a weight of 833 N dives from a boat with a mass of 375kg. If he leaves the boat with a velocity of 2.25m/s [W] after accelerating for 0.50s, what will be the final velocity of the boat?

The attempt at a solution
ad=Vfd-Vid/t
=2.25-0/0.5
ad=4.5m/s^2 [W]

Fg=mdg
833=m(9.8)
md=85kg

Fnet=mdad
Fg+Fd=mdad
833+Fd=-450.5

Fb+Fd=0
Fb= -Fd
Fb= 450.5N

Fb=mbab
450.5=(375)(ab)
ab=1.26m/s^2

ab=Vfb-Vib/t
1.26=Vf-0/0.50
Vf= 0.63m/s

Is this correct?
 
Physics news on Phys.org
No, that's not correct. You should ignore gravitational force here. Assume the diver leaves the boat with a horizontal velocity of 2.25 m/s. The only thing for which you need gravitational acceleration is the relationship between weight and mass.
 
D H said:
No, that's not correct. You should ignore gravitational force here. Assume the diver leaves the boat with a horizontal velocity of 2.25 m/s. The only thing for which you need gravitational acceleration is the relationship between weight and mass.

Fd=mdad
=(85)(4.5)
Fd=382.5 N

Fb= -382.5 N

Fb=mb(-ab)
-382.5=(375)(-ab)
(ab)= 1.02 m/s^2

ab= Vf-Vi/t
1.02=Vf-0/0.50
Vfb= 0.51m/s

So I would do this then?

Also, if the direction of the velocity was not given to be [W] then would I use the gravitational force?
 
Last edited:
You went from 382.5 N to 385.5 N. That's a mistake you carried through to your final answer.
 
D H said:
You went from 382.5 N to 385.5 N. That's a mistake you carried through to your final answer.

Thanks, fixed my solution. But if the diver was not traveling horizontally and the direction west was not stated. Then would gravitational force apply?
 
You have to know the direction. Velocity, acceleration, and force are vectors. Moreover, this is a 3rd law problem. The diver's interaction with the boat and with the Earth are two different interactions. You shouldn't mix them up.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Hot Threads

Recent Insights

Back
Top