A uniform rod of mass mr 118 g and length L 1000 cm is att

A uniform rod of mass

m_r = 118 g

and length

L = 100.0 cm

is attached to the wall with a pin as shown. Cords are attached to the rod at the

r₁ = 10.0 cm and r₂ = 90.0 cm

mark, passed over pulleys, and masses of

m₁ = 256 g and m₂ = 147 g

are attached. Your TA asks you to determine the following.

(a) The position r₃ on the rod where you would suspend a mass

m₃ = 200 g

in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle _F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative).

F_p =

_F =

(b) Let\'s now remove the mass m₃ and determine the new mass m₄ you would need to suspend from the rod at the position

r₄ = 20.0 cm

in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Express the direction of the force the pin exerts on the rod as the angle _F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative).

_F =
    (c) Let\'s now remove the mass m₄ and determine the mass m₅ you would suspend from the rod in order to have a situation such that the pin does not exert a force on the rod and the location r₅ from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position.

Solution

According to the given problem,

a) Using conservation of energy principal,

0.200*r₃*g + 0.50*0.118*g = 0.256*0.1*g + 0.147*0.9*g or

r₃ = 5*[0.0256+0.1341 -0.059] = 0.5035m

Force = 0.200*9.81 = 1.962 N at an angle of 270° from horizontal

b) Same as above

m₄*0.20*g + 0.50*0.118*g = 0.256*0.1*g + 0.147*0.9*g or

m₄ = 5*[0.0256+0.1341 -0.059] = 0.5036 kg or 503.36 g

Force = 0.200*9.81 = 1.962 N at an angle of 270° from horizontal

c) As it be cannot understand! If mass does not exert any force then any mass will do, We have to find only the point from which it needs to be suspended. But such a point seems to be impossible as the torque exerted by it will be zero and the rod cannot remain horizontal.