Angle Between 2 Forces of 14lb & 30lb Resulting in 20lb

  • Thread starter Thread starter dandogg
  • Start date Start date
AI Thread Summary
To find the angle between two forces of 14lb and 30lb resulting in a 20lb force, one can use the Law of Cosines. First, create a diagram with the two forces and the resultant force, forming two triangles. The sides of the triangle will be 14, 20, and 30, with the angle between the 14lb and 30lb forces being the focus. The cosine law will help determine this angle, which can be either acute or obtuse. Proper vector addition is essential for accurate calculations in this scenario.
dandogg
Messages
2
Reaction score
0
hi i was wondering if there was a formula to find the angle between two forces
i.e. two forces of 14lb and 30lb act on a body forming an obtuse angle with each other having a resulting force of 20lbs
 
Physics news on Phys.org
Its a trig problem. Draw a picture of the two forces and the resultant force. create two triangles, and label all the sides. then use law of cosines to find one angle, and then the other angle. sum the two angles for the total angle between the two forces.
 
Theres two ways you can add the two forces to get a net force of 20lbs. One is acute and one is that same angle + 90 deg (obtuse).

You'll want to use vector addition either way.
 
so if i make a triangle, the largest force of 30would be the hypotenuse and the magnitude and the
other force would be the other two sides?
 
Last edited:
No, strictly speaking, the term "hypotenuse" only applies to right triangles. It is true, of course, that in your triangle, the sides will have "lengths" 14, 20, and 30. Since the 14 and 30 pound forces are the ones adding, you want the angle between those sides.
Use the cosine law to find that angle.
 
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