3 forces act in equilibrium, calculate their angles

AI Thread Summary
To calculate the angles between three forces in equilibrium (12N, 5N, and 13N), it is essential to establish that the resultant force is zero. The problem can be approached by constructing a scalene triangle and resolving the forces along the x and y axes to create equations. Using the cosine rule is a valid method, but it requires careful consideration of the triangle formed by the forces. The longest vector should be placed along the negative x-axis to simplify calculations. Understanding the relationship between the triangle's angles and the vector angles is crucial for accurate results.
alanm
Messages
4
Reaction score
0
I have 3 forces, I assume that the resultant is equilibrium, that is just a guess.

a - 12N
b - 5N
c - 13N

how do i calculate the angles between the forces.

I think I have to use the cosine rule but am not really sure. I have a book Physics made Simple but it doesn't really help

Alan
 
Physics news on Phys.org
If the forces are in equilibrium, then the resultant is 0. And this is the correct conlusion and not a guess.Right?
 
hi,
the bits I need are the angles between each set of forces. I am just not clear on how to do this.

alan
 
Hi Alan,

If the result is not equilibrium then this problem is currently unsolvable, so assume it is. This is evident by the fact you have 3 unknown variables, the angle between each force, and you require at least 3 equations to solve this.

The first two equations you may construct by drawing the problem as a scalene triangle with 3 unknowns, then setting one of the forces along the x-axis (you could equally use the y axis). Then resolve the forces along the x and y-axis giving the first 2 equations:

note: you will need to define a direction for each force. I have assumed directions, did they give you this information?

Along x: 1) ...

Along y: 2) ...

You can then use the cosine rule, and or a+b+c=180 (if a,b,c are the 3 angles). Let me know how you get on
 
There are several ways to approach the problem. One is to construct a set of simultaneous equations involving the vector components and solve for them, then determine the angles from there. A simple approach is to put the longest vector on the minus x-axis (so it has no y-components, just an x-component), then find the orientation of the two other vectors that have zero sum y components and x-components that cancel the first vector. Four equations in four unknowns.

Another approach is to use, as you've surmised, the cosine rule. A triangle constructed from sides whose lengths are the vector lengths will have angles that are related to (but not the same as!) those between the vectors. Subtract the angles you find from pi to determine the inter-vector angles. (If you're interested, it has to do with the exterior angles of the triangle and their relationships to the vector angles). This is a pretty good approach!
 
thanks, I came up with the cosine rule but struggled to get there.

alan
 
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...
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}...
Back
Top