A 40-N crate rests on a rough horizontal floor. A 12-N horizonatl force is then applied to it. If the coefficients of friction are s = 0.5 and k = 0.4, the magnitude of the frictional force on the crate is:(adsbygoogle = window.adsbygoogle || []).push({});

what i did was :

40*0.5=20 , 40*0.4=16 20+16=36 40/12=3.3 36/3=12

how to solve this problem? could someone help?

thank you.

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# Homework Help: Friction - could you help me?

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