Ok. So coordinates for each point would be in relation to (0.2,0) If I do it how you suggest:
T1(0m) + T2(1.6m) = (45kg)(9.81m/s2)(0.91m)
T2 = 250.8N
Then can I use my earlier equation to solve for T1?
T1(0.2m) + T2(1.8m) = (45kg)(9.81m/s2)(1.11m)
0.2T1 + (250.8N)(1.8m) =...
Homework Statement
A 30Kg neon sign is suspended by two cables, as shown. Three neighbourhood cats (5.0 Kg each) find the sign a comfortable place. Calculate the tension in each cable with the cats are in the positions shown.
Homework Equations
xCM = m1x1 + m2x2 ...
Homework Statement
A uniform vertical beam of mass 40kg is acted on by a horizontal force of 520N at its top and is held, in the vertical position, by a cable as shown in the attached picture.
a) Calculate the tension in the cable
b) Determine the reaction forces acting on the beam by the...
1) Would that be at the end of the rod, the opposite end of the pivot?
2) if the point of application is the end, then it would be 1.0 m and the radius should be 1.0m, right?
Ok, so:
τ = Iα
τ = 1/3ml2α
τ = rF
F = mg
τ = rmg
rmg = 1/3ml2α
(1.0m)(12kg)(9.81m/s2) = (1/3)(12kg)(1m)2α
α = 29.4rad/s2
This seems rather high...
would the radius of the rod be its length because it is pivoting on its end? I think i did something wrong... thanks for the help
Homework Statement
Consider a uniform rod of mass 12kg and length 1.0m. At its end, the rod is attached to a fixed, friction-free pivot. Initially the rod is balanced vertically above the pivot and begins to fall (from rest) as shown in the diagram. Determine,
a) the angular acceleration of...