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Question Number 227561 by mr W last updated on 08/Feb/26

Commented by mr W last updated on 08/Feb/26

question as Q227521, all blocks have same size and roughness, but different weights. block no. 1 has the weight 1N. block no. 2 has the weight 2N. ... block no. 100 has the weight 100N. find the friction force between block no. 27 and block no. 28.

$${question}\:{as}\:{Q}\mathrm{227521},\:{all}\:{blocks} \\ $$$${have}\:{same}\:{size}\:{and}\:{roughness},\:{but} \\ $$$${different}\:{weights}. \\ $$$${block}\:{no}.\:\mathrm{1}\:{has}\:{the}\:{weight}\:\mathrm{1}{N}. \\ $$$${block}\:{no}.\:\mathrm{2}\:{has}\:{the}\:{weight}\:\mathrm{2}{N}. \\ $$$$... \\ $$$${block}\:{no}.\:\mathrm{100}\:{has}\:{the}\:{weight}\:\mathrm{100}{N}. \\ $$$${find}\:{the}\:{friction}\:{force}\:{between} \\ $$$${block}\:{no}.\:\mathrm{27}\:{and}\:{block}\:{no}.\:\mathrm{28}. \\ $$

Answered by mr W last updated on 08/Feb/26

Commented by mr W last updated on 09/Feb/26

a=width of the block G_i =weight of block No. i=i N f_0 ×100a=Σ_(i=1) ^(100) G_i ×(100a−ia+(a/2)) f_0 =(1/(100a)) Σ_(i=1) ^(100) (100a−ia+(a/2))i =(1/(100))Σ_(i=1) ^(100) (100−i+(1/2))i =(((100+(1/2))×((100×101)/2)−((100×101×201)/6))/(100)) =((6767)/4) N f_k =friction force between block i and i+1 f_0 +f_k =Σ_(i=1) ^k G_i =Σ_(i=1) ^k i f_k =f_0 −Σ_(i=1) ^k i=((6767)/4)−((k(k+1))/2) f_(27) =((6767)/4)−((27×28)/2)=((5255)/4) N

$${a}={width}\:{of}\:{the}\:{block} \\ $$$${G}_{{i}} ={weight}\:{of}\:{block}\:{No}.\:{i}={i}\:{N} \\ $$$${f}_{\mathrm{0}} ×\mathrm{100}{a}=\underset{{i}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}{G}_{{i}} ×\left(\mathrm{100}{a}−{ia}+\frac{{a}}{\mathrm{2}}\right) \\ $$$${f}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{100}{a}}\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\left(\mathrm{100}{a}−{ia}+\frac{{a}}{\mathrm{2}}\right){i} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{100}}\underset{{i}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\left(\mathrm{100}−{i}+\frac{\mathrm{1}}{\mathrm{2}}\right){i} \\ $$$$\:\:\:\:=\frac{\left(\mathrm{100}+\frac{\mathrm{1}}{\mathrm{2}}\right)×\frac{\mathrm{100}×\mathrm{101}}{\mathrm{2}}−\frac{\mathrm{100}×\mathrm{101}×\mathrm{201}}{\mathrm{6}}}{\mathrm{100}} \\ $$$$\:\:\:\:=\frac{\mathrm{6767}}{\mathrm{4}}\:{N} \\ $$$${f}_{{k}} ={friction}\:{force}\:{between}\:{block}\:{i}\:{and}\:{i}+\mathrm{1} \\ $$$${f}_{\mathrm{0}} +{f}_{{k}} =\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}{G}_{{i}} =\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}{i} \\ $$$${f}_{{k}} ={f}_{\mathrm{0}} −\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}{i}=\frac{\mathrm{6767}}{\mathrm{4}}−\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${f}_{\mathrm{27}} =\frac{\mathrm{6767}}{\mathrm{4}}−\frac{\mathrm{27}×\mathrm{28}}{\mathrm{2}}=\frac{\mathrm{5255}}{\mathrm{4}}\:{N} \\ $$

Commented by mr W last updated on 09/Feb/26

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