A 2.0 kg block is sitting on an incline at an angle of 52.1 above the horizontal. The magnitude of the static friction force, in N, needed to keep the block stationary, is

the free body diagram of block sitting on an incline at an angle 0= 52 -10 above horizontal is N m - mgsind mg Coto in free body diagram N is normal force on block, and fs is Static Brickon force. from free body diagram we can see that the block will move down the incline with force . masino then static friction force will act opposite to the motion of block. To keep the block stationany the friction force fs should be equal to mg sino. So Is = mgsind given, m = 2.0 kg , 0= 52.10 g = 9.8 m So , fs = ( 2 . 0 k g ) x ( 9 . 8 ! 2 1 x Sin (52. 10) fs = 15. 466 N fs = 15 . 47 N