What is the magnitude and direction of the second force vector in this scenario?

  • Thread starter Thread starter LuvIz4ever
  • Start date Start date
AI Thread Summary
The discussion revolves around a physics problem involving two force vectors, F1 and F2, where F1 is 200 N east and the resultant vector is 400 N along the east-west line. The confusion arises regarding the direction of F2, with some suggesting it could be described as having a north component. However, it is clarified that F2 must be 600 N west, as adding 200 N east and 600 N west results in a net force of 400 N west. The participants agree that the answer book's reference to "north" is incorrect, as the vectors only lie along the east-west axis. Ultimately, the correct interpretation is that F2 is simply 600 N west, without any northward component.
LuvIz4ever
Messages
6
Reaction score
0
I can't figur this out!

Now i got the answer to this problem but i am so confused at this one thing... how is it that the answer is 200 N of East and 600 N of west


the question is

A force vector F1 points due east with a magnitude of 200 Newtons and a second force F2 is added to F1. The resultant of the two vectors has a magnitude of 400 Newtons and points along the East and west line. Find the magnitude and direction of F2 note that there are 2 answers.


My question is I know the answers 2oo east and 6oo west but why north of east or north of west.
 
Physics news on Phys.org
LuvIz4ever said:
how is it that the answer is 200 N of East and 600 N of west
I suspect that the N stands for Newtons, not north.
 
Thanks But...

Actually in the answer book it tells me it is North... I think it is weird though becuase of the fact that it has nothing about North.. in the question
 
The answer in your book must be wrong. You have an initial vector pointing along the east-west line. If you add to it a vector with any component at all in the north-south direction, the resultant cannot possibly be entirely along the east-west line. I suspect the Doc is absolutely right.
 
... I figured that... thanks! sooooo! Much!
 
simply put, there is 200 going "right" and 600 going "left" add 200 + (-600) leads to a net of 400, and this 400 is along EW or as most would say x-axis..

yw
 

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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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 »

Similar threads

Plane Traveling against the wind
Replies
4
Views
1K
Direction of the magnetic needle
Replies
11
Views
2K
I am calculating the forces on String cheese with net force = 0
Replies
13
Views
2K
Two forces acting on an object given in vectors - SOLVED
Replies
1
Views
1K
Doubt about the 2nd position vector direction in a particle's movement
Replies
4
Views
1K
Equilibrium of Forces and Torques on Sawhorses Supporting a Person on a Board
Replies
6
Views
2K
Friction problem of a block sandwiched between 2 surfaces
Replies
6
Views
1K
Three equal forces applied to a rectangle, find net torque direction?
Replies
30
Views
4K
Coriolis Force on a Race Car at 45 Degrees North
Replies
3
Views
2K
Displacement -0.25i + 14.7j from Origin: Final Displacement
Replies
4
Views
2K

Hot Threads

Recent Insights

Back
Top