Refer to the diagram here (a side view, not to scale), where everything is at re
ID: 1526270 • Letter: R
Question
Refer to the diagram here (a side view, not to scale), where everything is at rest. A horizontal rod of length L is pinned at its left end to a vertical wall. Let the location of that pin cable define the origin of a set of x-y axes (all coordinates in m). At the point (0, h), a massless cable is attached to the wall The other end of the cable is fastened to the rod at (L/2, 0). A block is suspended from the rod by a massless wire at (d, 0). The block and the rod have equal masses, but the rod block has a non-uniform linear mass density: lambda(x) = C - Bx^2/3 Here are the known data: L = 7.31 m h = 2.46 m d = 5.14 m C = 2.76 kg/m B = 0.538 kg m^-5/3 g = 9.80 m/s^2 Calculate the magnitude of the tension in the cable Calculate the force (magnitude and direction) exerted on the rod by the pin at its left end. Now suppose that the rod is held at rest (still horizontal) while the cable, block and wire are removed then the rod is released. At what speed would its midpoint strike the wall?Explanation / Answer
Ans:-
a.By using Newton’s second law
on x axis
Fpin = Tx...........1
On y axis
Ty = Fwire +mg
Apply law on box
Fwire = Fg ..............2
Ty = Fg + mg ............3
Apply torque on rod
Fpin *0 + Ty*L/2 –mg*L/2 -Fwire* (d) = 0.............4
= tan^-1(2.46/3.66) = 33.94 = 34deg
= C-Bx^2/3 = 2.76 – 0.538*(7.31)^2/3 = 0.72kg/m
= m/L
m= *L = 0.72*7.31 = 5.264kg
put value in equation 4
T *Sin34*3.66 – 5.26*9.8*3.66 – 5.26*9.8*5.14 = 0
T = 453.6224/sin34*3.66 = 221.64N
b. Fpin = Tx = T cos 34 = 221.64*cos34 = 183N +ve X direction