A thin cylindrical rod of uniform mass density, rests against a smooth wall as shown in figure.
of friction between ground and rod (if the minimum angle that the rod can make with the horizontal is theta)
Answers
coefficient of friction between ground and rod is (1/2)cosθ
we have to draw free body diagram of given situation.
see figure,
here N1 is normal reaction acting between ground and rod and N2 is the normal reaction acting between wall and rod,
friction, f = μN1 [ from figure ]
weight of rod = mg [ vertically downward from middle position of rod]
now, N2 = fr = μN1 .....(1)
Mg = N1 ......(2)
torque at point A = 0 [ rod doesn't rotate ]
torque due to N1 + torque due to weight + torque due to N2 + torque due to friction = 0
⇒0 + Mgcosθ × l/2 + 0 - (N2sinθ) × l = 0
⇒Mgcosθ = 2N2sinθ.......(3)
from equations (1), (2) and (3),
we get, (Mgcotθ)/2 = μMg
⇒ μ = (1/2)cotθ
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coefficient of friction between ground and rod is (1/2)cosθ
we have to draw free body diagram of given situation.
see figure,
here N1 is normal reaction acting between ground and rod and N2 is the normal reaction acting between wall and rod,
friction, f = μN1 [ from figure ]
weight of rod = mg [ vertically downward from middle position of rod]
now, N2 = fr = μN1 .....(1)
Mg = N1 ......(2)
torque at point A = 0 [ rod doesn't rotate ]
torque due to N1 + torque due to weight + torque due to N2 + torque due to friction = 0
⇒0 + Mgcosθ × l/2 + 0 - (N2sinθ) × l = 0
⇒Mgcosθ = 2N2sinθ.......(3)
from equations (1), (2) and (3),
we get, (Mgcotθ)/2 = μMg
⇒ μ = (1/2)cotθ
read more similar questions :A rod of mass m and length l is kept on a smooth wedge of mass M as shown in the figure . If the system is released when...
brainly.in/question/737703
A uniform rod of mass m and length l is placed horizontally on a smooth horizontal surface an impulse p is applie to end...
brainly.in/question/6865092