Investigate Newtons Laws. Intro Physics

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 2K views
Kmcquiggan
Messages
29
Reaction score
1
Homework Statement
Two tugboats are pulling on a large log, as shown in figure. The log has a mass of 250kg and is initially at rest. How far has tug boat gone after 10s?
Relevant Equations
v_2=v_1+a∆, c= √(a^2+b^2-2abcos°), sine law=sinA/a=sinB/b=sinC/c, a=F_net/m
244483
244484
 
Physics news on Phys.org
yes the question is in the question area as well as in the first picture and my attempt is in the second image which is a screen shot of my laptop screen.
 
Kmcquiggan said:
Problem Statement: Two tugboats are pulling on a large log, as shown in figure. The log has a mass of 250kg and is initially at rest. How far has tug boat gone after 10s?
Relevant Equations: v_2=v_1+a∆, c= √(a^2+b^2-2abcos°), sine law=sinA/a=sinB/b=sinC/c, a=F_net/m
What is your c= √(a^2+b^2-2abcos°)? You need the sum of the two forces, but you calculated the difference.
 
Looks like you tried to use the law of cosines to find the net force: Better recheck your arithmetic.

Once you find the acceleration, how would you find the distance traveled?
 
°I went back and checked and see that I made a calculation wrong C= 1163.72 round to 1164. and I then reput the info into sine law and got Fnet= 20°-10° which left [W 10 ° N] than I figured a = Fnet/m = 1164/250=4.656
I used the distance formula which is:
d=vi8t + 1/2*a*t^2 = d= 0+(.5)(4.66)(100) =233m
So therefore the tg boats would be 233m after 10 sec is this getting closer?
 
Kmcquiggan said:
I made a calculation wrong C= 1163.72 round to 1164.
That looks better.
Kmcquiggan said:
I then reput the info into sine law and got Fnet= 20°-10° which left [W 10 ° N]
?? Why are you using the sine law?
Kmcquiggan said:
than I figured a = Fnet/m = 1164/250=4.656
Looks OK. As does the rest of your work.