Tension of wires holding a suspended sign in equilibrium

AI Thread Summary
To find the tension in the wires holding a 15kg picture, the problem involves breaking the forces into vertical and horizontal components based on the angles of 220° and 320°. The user initially set up the equation for vertical forces but struggled with substituting values correctly. It's important to ensure the correct signs for the components based on the quadrant of the angles. Creating a sketch can help clarify the relationships between the angles and the forces involved. Accurate component analysis is crucial for solving for the tension in each wire.
tigerlily5709
Messages
1
Reaction score
0
1.Hey guys, I've been struggling with this one for an hour and it SHOULD NOT be this hard for me.

A 15kg picture hangs from a hook on the wall. What is the tension in each wire if one is at a cartesian angle of 220° and one is at a cartesian angle of 320°?



2.

So my physics 20 teacher told me to break it into vertical and horizontal components. I solved for one of my tensions and then tried to substitute into my second equation...that's when things get scary. Please help!



3. I tried:

0N = -147.15N + Ft1 (sin 220°) + Ft2 (sin 320°)

I was trying to solve for Ft1 first.
 
Physics news on Phys.org
tigerlily5709 said:
1.Hey guys, I've been struggling with this one for an hour and it SHOULD NOT be this hard for me.

A 15kg picture hangs from a hook on the wall. What is the tension in each wire if one is at a cartesian angle of 220° and one is at a cartesian angle of 320°?



2.

So my physics 20 teacher told me to break it into vertical and horizontal components. I solved for one of my tensions and then tried to substitute into my second equation...that's when things get scary. Please help!



3. I tried:

0N = -147.15N + Ft1 (sin 220°) + Ft2 (sin 320°)

I was trying to solve for Ft1 first.

You should check to see what quadrants your angles are in, and so make sure that the components you're extracting via sin and cos have the signs that you're expecting.

It's often expedient to make a sketch and use the given angles to label other angles that are more conveniently associated with the "usual" analysis steps.
 
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