Answer:
1) [tex] W_{object} = 400 N [/tex]
2) x = 0.28 m from cable A.
Explanation:
1 ) Let's use the first Newton to find the , because bar is in equilibrium.
[tex] \sum F_{Tot} = 0 [/tex]
In this case we just have y-direction forces.
[tex] \sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0 [/tex]
Now, let's solve the equation for W(object).
[tex] W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N [/tex]
2 ) To find the position of the heaviest weight we need to use the torque definition.
[tex] \sum \tau = 0 [/tex]
The total torque is evaluated in the axes of the object.
Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.
First let's find the positions from each force to the P point.
[tex] L = 1.50 m [/tex] ; total length of the bar.
[tex] D_{AP} = x [/tex] ; distance between Tension A and P point.
[tex] D_{BP} = L-x [/tex] ; distance between Tension B and P point.
[tex] D_{W_{bar}P} = \frac{L}{2}-x [/tex] ; distance between weight of the bar (middle of the bar) and P point.
Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.
[tex] \sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0 [/tex]
Finally we just need to solve it for x.
[tex] x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}} [/tex]
[tex] x = 0.28 m [/tex]
So the distance is x = 0.28 m from cable A.
Hope it helps!
Have a nice day! :)
Answer:
Weight = 400 N
d = 0.28m
Explanation:
∑Fy = 0 gives Ta + Tb - W - Wbar = 0.
W=Ta + Tb - Wbar = 550N + 300N - 450N = 400N
To find distance you need to use the net torque formula.
∑τ = 0 gives Tb(d) - W * d - Wbar * ([tex]\frac{d}{2}[/tex])
τ = Force * distance
d = [tex]\frac{Tb*(1.5) - Wbar(0.75)}{W}[/tex]
d = 0.28m
