A mass of 6 kg is placed on a horizontal table with friction. A rope is tied to

Question

A mass of 6 kg is placed on a horizontal table with friction. A rope is tied to it and it is slung over a pulley and the other end of the rope is attached to a hanging mass of 16 kg. The hanging mass is great enough such that it overcomes the static friction of the mass on the table and the hanging mass accelerates downward at 2.1 m/s2. If the coefficient of static friction is 4 times larger than the coefficient of kinetic friction, and if the hanging mass is changed, what is the minimum the hanging mass needs to be to begin to move the mass on the table, in kg?

Explanation / Answer

accleration= 2.1 m/s^2

frictional force=mass*accleration=6* 2.1 = 12.6 N

accleration due to gravity=g=9.8

the coefficient of the kinetic friction=

frictional force=mass*accleration due to gravity*thecoefficient of the kinetic
friction =16 *9.8* u

=12.6/(16* 9.8)

u = 0.08

the coefficient of static frictionis=s=4* 0.08 = 0.321

the minimum horizontal force needed in order to move thebox

again =mass*g*s = 16 * 9.8 * 0.321 = 50.32 N

minimum mass 18.8/6 = 8.38 kgs

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