Force at Equilibrium: How to Solve X and Y Components of Tension

  • Thread starter Thread starter GOPgabe
  • Start date Start date
  • Tags Tags
    Equilibrium Force
AI Thread Summary
To solve for the tension components in a system of ropes, the X and Y components must be balanced, leading to two equations with two unknowns. The discussion highlights the importance of understanding that at the junction where the ropes meet, the sum of horizontal and vertical forces must equal zero. A participant initially miscalculated the relationship between the forces and sought clarification on whether to combine vertical components to determine the weight of Fw1. It was clarified that the relevant system is the knot where the ropes converge, emphasizing the need to consider all acting forces. Understanding these principles is crucial for accurately solving tension problems in equilibrium.
GOPgabe
Messages
10
Reaction score
0

Homework Statement


I have an actual image which is quite better than my poor descriptive skills

PhysProblem.png


Homework Equations


The X and Y components of each tension must balance.

The Attempt at a Solution



I thought if F3 weighs 200N then that mean the X component of Fw1 must equal
Fw1 tan(35)=200. Clearly I was wrong based on the answer. How would I proceed from here? Any help would be much appreciated.
 
Physics news on Phys.org
At the point where the three ropes meet, the sum of all the horizontal forces is zero and the sum of all the vertical forces is zero. This gives you two equations and you have two unknowns.
 
kuruman said:
At the point where the three ropes meet, the sum of all the horizontal forces is zero and the sum of all the vertical forces is zero. This gives you two equations and you have two unknowns.

I think I got it now, just one question. Are you supposed to add the vertical components of both of the systems in order to get the weight of Fw1?
 
GOPgabe said:
I think I got it now, just one question. Are you supposed to add the vertical components of both of the systems in order to get the weight of Fw1?
I am not sure what you mean by "both systems". Your system in this case is the knot where all three ropes are tied together. There are three forces acting on it, and the sum of these forces is zero.
 
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