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# There is a table, and a napkin on it. The napkin has a mass m n=92. On the napkin, there is a tea cup with mass m c=199.

There is a table, and a napkin on it. The napkin has a mass mn=92.8 g. On the napkin, there is a tea cup with mass mc=199.2 g. What is the minimum force required for someone to pull the napkin out from under the tea cup? Assume that between the table and the napkin the friction is µk= 0.293, while the coefficient of static friction between the tea cup and the napkin is µs=0.637. Express your answer in Newtons.

Using Newton's Second law of motion F = ma ΣFxNapkin=F-frC-frT (where C represents the cup and T is Tue... View the full answer

The way to answer this question is ... View the full answer

• Do you have any other suggestions. I keep getting the wrong answer.
• hossu
• Oct 03, 2018 at 5:56pm
• Is the answer = 2.08 N ? If yes, then I will give you explanation.
• ShubhamVyas
• Oct 03, 2018 at 11:20pm
• I got a another answer and may be this is right. Since, corrected equation of motion of napkin is, " mn*a = F - uk*(mn + mc)*g + us*mc*g " and equation of motion of cup is, " mc*a = us*mc*g " Now put value of (a) from second equation to first equation, then we have net equation as , " mn*us*g = F - uk(mn + mc)*g + us*mc*g "
• ShubhamVyas
• Oct 04, 2018 at 12:31pm
• After solving above equation I got an answer :- F = 0.174 N
• ShubhamVyas
• Oct 04, 2018 at 12:31pm
• If my answer is correct, please let me know.
• ShubhamVyas
• Oct 04, 2018 at 12:32pm

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