Calculating Resultant Force: Comparing Methods and Identifying Errors

AI Thread Summary
The discussion focuses on calculating the resultant force using two different methods: the law of cosines and vector component breakdown. The initial calculation using the law of cosines yields a resultant force of 49N, while the component method results in 55N. Participants suggest that breaking down each force vector into x and y components and then applying the Pythagorean theorem is the most effective approach. The discrepancy between the two methods raises questions about the accuracy of the initial calculation. Ultimately, the consensus leans towards the vector component method as the more reliable technique for determining the resultant force.
vipertongn
Messages
97
Reaction score
0
Ok so I have something like this...not drawn to scale

http://i53.tinypic.com/303l5k6.gif

I can see that with the law of cos I can get 49N

However...with this other method where you set Sum of F=0

With sqrt(Fx^2+Fy^2)=R

sqrt((40*cos(20)+20*cos(30))^2+(40*sin(20)-20*sin(30))^2)

I get 55N

Is there soemthing wrong? I'm pretty sure the first answer is correct...but I don't know why this one isnt...
 
Physics news on Phys.org
The second one looks right, so I'm guessing the first is wrong... what do you mean using the law of cos?

I think the best way is to just break down each force vector into its x and y components; add those to find the resultant, then use the Pythagorean theorem to find the magnitude of the resultant (that's effectively what you did with your second method).
 

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 »

Similar threads

Vectors Directions: Where is this Resultant Vector Pointing?
Replies
14
Views
1K
Statics problem on the resolution of forces
Replies
18
Views
2K
Pressure question: Weight of a car calculated from tyre contact patch area
Replies
3
Views
860
Where on a rigid body is a resultant force applied?
Replies
11
Views
1K
Is My Approach to Calculating Relative Velocity Correct?
Replies
20
Views
2K
To find the resultant of forces on a lamina
Replies
5
Views
3K
Seemingly simple equilibrium problem - force balance
Replies
6
Views
2K
Find moment arm of resultant force of two forces applied at two points
Replies
3
Views
958
Block on top of an inclined plane, with a rough surface, that moves with constant acceleration
Replies
41
Views
4K
Finding the Magnetic force on the coil
Replies
3
Views
6K

Hot Threads

Recent Insights

Back
Top