A chain of mass 10 kg and 10 m long is resting on a rough surface horizontal. Coefficient of friction is 0.2.

• Mass of chain (m) = 10 kg


• Length of chain (l) = 10 m


• Coefficient of friction (u) = 0.2

Weight of chain is given by:
W = mg
where,
m = mass of chain = 10 kg
g = acceleration due to gravity = 9.8 m/s^2
Therefore,
W = 10 kg x 9.8 m/s^2 = 98 N


The chain is at rest and hence the tension in the chain is equal to its weight. Hence,
T = W = 98 N


The frictional force acting on the chain is given by:
f = uN
where,
u = coefficient of friction = 0.2
N = normal force
The normal force is the force exerted by the surface on the chain perpendicular to the surface. In this case, the chain is resting on a horizontal surface and hence the normal force is equal to the weight of the chain i.e. N = W = 98 N
Therefore,
f = 0.2 x 98 N = 19.6 N


For the chain to be in equilibrium, the net force acting on it should be zero.
Net force = T - f
= 98 N - 19.6 N
= 78.4 N
Since the net force is non-zero, the chain is not in equilibrium and will start moving.


The acceleration of the chain is given by:
a = (T - f)/m
= (98 N - 19.6 N)/10 kg
= 7.84 m/s^2


The chain of mass 10 kg and 10 m long resting on a rough surface with a coefficient of friction of 0.2 will start moving with an acceleration of 7.84 m/s^2.

As it rest on a surface hence the normal force =mg =100
f = 0.2 ×100 = 20 n
Also never mind with length !!