1frictional forcemax =uN (u=.2,N=mg cos 300)
=2√3 N
& component of force actin downwards=mg sin 300=10 N.
wich is already>2√3N. since no other force is actin upwards ......so no extra force is required to overcome the friction.
A BODY OF MASS 2 KG IS LYING ON A ROUGH INCLINED PLANE OF INCLINATION 30 DEGREES.FIND MAGNITUDE OF FORCE PARALLEL TO INCLINE NEEDED TO MOVE BLOCK DOWN THE INCLINE.COEFFICIENT OF STATIC FRICTION=0.2
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7 Answers
11tan 30 = 1/√3 = 0.58
This should be the friction at 30° but it is 0.2 therefore No Force will be required
11virang answer is zero....we r rite......john msged me in chat box......but who is brinda??
