disruptors
Nov18-04, 08:57 PM
A horizontal force of 12 N pushes a 0.5 Kg block against a vertical wall. The block is initially at rest, If static and kinetic coefficients are 0.6 and 0.8 respectively what is the friction force?
Basically the fbd has Force of 12 N across the x component(+) and the Normal force in the opposite (-ve)... The y component is positive facing downwards thus Fnety=mg-frictionforce=ma(y)... Friction force is equal to coefficient of static x Normal... static is 0.6 and Normal force is 12 N, which gives 7.2 N, right?
(i am doubting myself because some other ppl didnt get that)
Thanks
Basically the fbd has Force of 12 N across the x component(+) and the Normal force in the opposite (-ve)... The y component is positive facing downwards thus Fnety=mg-frictionforce=ma(y)... Friction force is equal to coefficient of static x Normal... static is 0.6 and Normal force is 12 N, which gives 7.2 N, right?
(i am doubting myself because some other ppl didnt get that)
Thanks