Menu Close

Question-227521




Question Number 227521 by mr W last updated on 07/Feb/26
Commented by mr W last updated on 08/Feb/26
As shown in the figure, 100 wooden blocks, each with weight 5 N, are clamped by two forces of magnitude F from the left and right. All contact surfaces have the same roughness. What is the friction force between block No. 27 and block No. 28?
Commented by MrAjder last updated on 08/Feb/26
It should be 250N - it's not hard for readers to verify on their own
Commented by MrAjder last updated on 08/Feb/26
yes
Commented by mr W last updated on 08/Feb/26
the question is to find the friction  force between block no. 27 and block  no. 28. do you want to say that this  force is 250 N?
$${the}\:{question}\:{is}\:{to}\:{find}\:{the}\:{friction} \\ $$$${force}\:{between}\:{block}\:{no}.\:\mathrm{27}\:{and}\:{block} \\ $$$${no}.\:\mathrm{28}.\:{do}\:{you}\:{want}\:{to}\:{say}\:{that}\:{this} \\ $$$${force}\:{is}\:\mathrm{250}\:{N}? \\ $$
Commented by mr W last updated on 08/Feb/26
total weight of all blocks is 500 N.  since all blocks are identical, the  center of gravity of them lies in the  middle.  the friction force on the  left and on the right should be  equal. each of them is then  500/2=250N.  the friction force between block 1  and block 2 must be 5 N (=weight of  block 1) less than 250 N. etc.   the friction force between block 27   and 28 is 27×5N less than 250 N. so  the correct answer is  250−27×5=115 N.
$${total}\:{weight}\:{of}\:{all}\:{blocks}\:{is}\:\mathrm{500}\:{N}. \\ $$$${since}\:{all}\:{blocks}\:{are}\:{identical},\:{the} \\ $$$${center}\:{of}\:{gravity}\:{of}\:{them}\:{lies}\:{in}\:{the} \\ $$$${middle}.\:\:{the}\:{friction}\:{force}\:{on}\:{the} \\ $$$${left}\:{and}\:{on}\:{the}\:{right}\:{should}\:{be} \\ $$$${equal}.\:{each}\:{of}\:{them}\:{is}\:{then} \\ $$$$\mathrm{500}/\mathrm{2}=\mathrm{250}{N}. \\ $$$${the}\:{friction}\:{force}\:{between}\:{block}\:\mathrm{1} \\ $$$${and}\:{block}\:\mathrm{2}\:{must}\:{be}\:\mathrm{5}\:{N}\:\left(={weight}\:{of}\right. \\ $$$$\left.{block}\:\mathrm{1}\right)\:{less}\:{than}\:\mathrm{250}\:{N}.\:{etc}.\: \\ $$$${the}\:{friction}\:{force}\:{between}\:{block}\:\mathrm{27}\: \\ $$$${and}\:\mathrm{28}\:{is}\:\mathrm{27}×\mathrm{5}{N}\:{less}\:{than}\:\mathrm{250}\:{N}.\:{so} \\ $$$${the}\:{correct}\:{answer}\:{is} \\ $$$$\mathrm{250}−\mathrm{27}×\mathrm{5}=\mathrm{115}\:{N}. \\ $$
Commented by mr W last updated on 08/Feb/26
then wrong
$${then}\:{wrong} \\ $$
Commented by MrAjder last updated on 08/Feb/26
So, sir, do you already know the correct answer? Can you write down the process??
Commented by fantastic2 last updated on 08/Feb/26
why flagged as inappropiate
$${why}\:{flagged}\:{as}\:{inappropiate} \\ $$
Commented by mr W last updated on 08/Feb/26
or in other way:  the friction force between block 27  and 28 is the same as that between  block 73 and 74, due to symmetry.  they together carry the weight from  blocks no. 28 to 73.  ⇒f_(27−28) =f_(73−74) =((46×5)/2)=115 N
$${or}\:{in}\:{other}\:{way}: \\ $$$${the}\:{friction}\:{force}\:{between}\:{block}\:\mathrm{27} \\ $$$${and}\:\mathrm{28}\:{is}\:{the}\:{same}\:{as}\:{that}\:{between} \\ $$$${block}\:\mathrm{73}\:{and}\:\mathrm{74},\:{due}\:{to}\:{symmetry}. \\ $$$${they}\:{together}\:{carry}\:{the}\:{weight}\:{from} \\ $$$${blocks}\:{no}.\:\mathrm{28}\:{to}\:\mathrm{73}. \\ $$$$\Rightarrow{f}_{\mathrm{27}−\mathrm{28}} ={f}_{\mathrm{73}−\mathrm{74}} =\frac{\mathrm{46}×\mathrm{5}}{\mathrm{2}}=\mathrm{115}\:{N} \\ $$
Commented by mr W last updated on 08/Feb/26

Leave a Reply

Your email address will not be published. Required fields are marked *