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Friction
", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "
The coefficient of friction is $\\mu_s=\\var{mu_s}$ between all surfaces of contact. Knowing that $M_1=\\var{M1}$kg and $M_2=\\var{M2}$kg, determine the smallest force $\\mathbf{P}$ required to start the block $M_2$ moving if:
", "advice": "(a)
\nFree body: $\\var{M1}$-kg block
\n
$W_1=M_1 g=\\var{m1}\\times 9.81=\\var{m1*9.81}\\,\\text{N}$
\n$+\\!\\!\\uparrow\\sum F_y=0:\\quad N_1=W_1$
\n$F_1=\\mu_s N_1=\\var{mu_s}\\times\\var{m1*9.81}=\\var{mu_s*m1*9.81}\\,\\text{N}$
\n$\\overset{+}{\\rightarrow}\\sum F_x=0:\\quad T-F_1=0,\\quad T=F_1=\\var{mu_s*m1*9.81}\\,\\text{N}$
\n\n\nFree body: $\\var{M2}$-kg block
\n
$W_2=M_2g=\\var{M2}\\times 9.81=\\var{m2*9.81}\\,\\text{N}$
\n$+\\!\\!\\uparrow\\sum F_y=0: \\quad N_2-N_1-W_2=0,\\quad N_2=W_1+W_2=\\var{(m1+m2)*9.81}\\,\\text{N}$
\n$F_2=\\mu_s N_2=\\var{mu_s}\\times\\var{(m1+m2)*9.81}=\\var{mu_s*(m1+m2)*9.81}\\,\\text{N}$
\n$\\overset{+}{\\rightarrow}\\sum F_x=0: -P+F_1+F_2+T=0$
\n\\[P=F_1+F_2+T=\\var{P1}\\,\\text{N}\\quad\\blacktriangleleft\\]
\n\n(b)
\nBlocks move together
\n
$W=(M_1+M_2)g=\\var{(m1+m2)*9.81}\\,\\text{N}$
\n$\\overset{+}{\\rightarrow}\\sum F_x=0:\\quad P-\\mu_s N=0$
\n$P=\\mu_s N=\\mu_s W=\\var{mu_s}\\times\\var{(m1+m2)*9.81}=\\var{P2}\\,\\text{N}$
\n\\[P=\\var{P2}\\,\\text{N}\\quad\\blacktriangleleft\\]
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", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "
The coefficient of friction is $\\mu_s=\\var{mu_s}$ between all surfaces of contact. Knowing that $M_1=\\var{M1}$kg and $M_2=\\var{M2}$kg, determine the smallest force $\\mathbf{P}$ required to start the block $M_2$ moving if:
", "advice": "(a)
\nFree body: $\\var{M1}$-kg block
\n
$W_1=M_1 g=\\var{m1}\\times 9.81=\\var{m1*9.81}\\,\\text{N}$
\n$+\\!\\!\\uparrow\\sum F_y=0:\\quad N_1=W_1$
\n$F_1=\\mu_s N_1=\\var{mu_s}\\times\\var{m1*9.81}=\\var{mu_s*m1*9.81}\\,\\text{N}$
\n$\\overset{+}{\\rightarrow}\\sum F_x=0:\\quad T-F_1=0,\\quad T=F_1=\\var{mu_s*m1*9.81}\\,\\text{N}$
\n\n\nFree body: $\\var{M2}$-kg block
\n
$W_2=M_2g=\\var{M2}\\times 9.81=\\var{m2*9.81}\\,\\text{N}$
\n$+\\!\\!\\uparrow\\sum F_y=0: \\quad N_2-N_1-W_2=0,\\quad N_2=W_1+W_2=\\var{(m1+m2)*9.81}\\,\\text{N}$
\n$F_2=\\mu_s N_2=\\var{mu_s}\\times\\var{(m1+m2)*9.81}=\\var{mu_s*(m1+m2)*9.81}\\,\\text{N}$
\n$\\overset{+}{\\rightarrow}\\sum F_x=0: -P+F_1+F_2=0$
\n\\[P=F_1+F_2=\\var{P1}\\,\\text{N}\\quad\\blacktriangleleft\\]
\n\n(b)
\nBlocks move together (same as in Question 4)
\n\\[P=\\var{P2}\\,\\text{N}\\quad\\blacktriangleleft\\]
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", "minValue": "P2-1", "maxValue": "P2+1", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "3", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": true, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Polinomio p(x) con parametros t y s ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Considere el polinomio $p(x)$ con parametros $t$ y $s$. $ p(x) = \\simplify{{coef2_x3}x^3+s*x^2+{coef2_x}x+t}\\text{.}$
\nEl polinomio $p(x)$:
\nNos dicen que el polinomio:
\nprimeramente, sustituyendo $x = \\simplify{-{c}}$ sobre $p(x)$ obtenemos
\n\\begin{align}
p(\\simplify{-{c}}) &= \\simplify[all,!collectNumbers, fractionnumbers]{{coef2_x3*(-c)^3}+{(-c)^2}s+{coef2_x*(-c)}+t},\\\\
&= \\simplify[all,fractionnumbers]{{coef2_x3*(-c)^3}+{(-c)^2}s+{coef2_x*(-c)}+t}.
\\end{align}
pero, por el teorema del resto $p(\\simplify{-{c}}) = \\var{rem1}$ (usando la primera viñeta), entonces esto se convierte en
\n\\begin{align}
\\simplify[all,fractionnumbers]{{coef2_x3*(-{c})^3}+s*{(-{c})^2}+{coef2_x*(-{c})}+t} &= \\var{rem1},\\\\
\\simplify[all,fractionnumbers]{s*{x}+t} &= \\simplify[all,fractionnumbers]{{rem1}-{coef2_x3*(-{c})^3}-{coef2_x*(-{c})}}.
\\end{align}
Similarmente, sustituyendo $x = \\simplify{-{d}}$ sobre $p(x)$, obtenemos
\n\\begin{align}
p(\\simplify{-{d}}) &= \\simplify[all,!collectNumbers, fractionnumbers]{{coef2_x3*(-{d})^3}+{(-{d})^2}s+{coef2_x*(-{d})}+t},\\\\
&= \\simplify[all,fractionnumbers]{{coef2_x3*(-{d})^3}+{(-{d})^2}s+{coef2_x*(-{d})}+t}.
\\end{align}
pero, por el teorema del resto $p(\\simplify{-{d}}) = \\var{rem2}$ (usando la segunda viñeta), entonces esto se convierte en
\n\\begin{align}
\\simplify[all,fractionnumbers]{{coef2_x3*(-{d})^3}+s*{(-{d})^2}+{coef2_x*(-{d})}+t} &= \\var{rem2},\\\\
\\simplify[all,fractionnumbers]{s*{y}+t} &= \\simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}.
\\end{align}
Tenemos ahora dos ecuaciones simultaneas para $s$ y $t$:
\n\\begin{align}
\\simplify[all,fractionnumbers]{s*{x}+t} = \\simplify[all,fractionnumbers]{{rem1}-{coef2_x3*(-{c})^3}-{coef2_x*(-{c})}} \\\\
\\simplify[all,fractionnumbers]{s*{y}+t} = \\simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}
\\end{align}
A continuación, restamos la segunda ecuación de la primera ecuación.
\nEsto nos permite cancelar los términos que involucran $ t $ y nos da una ecuación solo en términos de $ s $, que luego podemos reorganizar para encontrar el valor de $ s $.
\nRestar las dos ecuaciones da:
\n\n\\[\\simplify{s*{(-{c})^2-(-{d})^2}} = \\simplify[all,fractionnumbers]{{rem1 - coef2_x3*(-c)^3-coef2_x*(-c)-rem2+coef2_x3*(-d)^3+coef2_x*(-d)}}.\\]
\nEntonces, podemos reorganizar esta ecuación para que
\n\\[s = \\simplify[all,fractionnumbers]{{rem1 - coef2_x3*(-c)^3-coef2_x*(-c)-rem2+coef2_x3*(-d)^3+coef2_x*(-d)}/{{(-c)^2-(-d)^2}}}.\\]
\nPodemos calcular $ t $ sustituyendo nuestro valor de $ s $ en una de nuestras ecuaciones simultáneas originales. Por ejemplo, usemos la ecuación
\n\\[\\simplify[all,fractionnumbers]{s*{(-{d})^2}+t} = \\simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}.\\]
\nSustituyendo nuestro valor de $ s $ en esta ecuación nos da
\n\\[
\\begin{align}
\\simplify[all,fractionnumbers,!noleadingMinus]{{numerator/denominator}+t} &= \\simplify[all,fractionnumbers]{{rem2-coef2_x3*(-d)^3-coef2_x*(-d)}},\\\\
t &= \\simplify[all,fractionnumbers]{{rem2-coef2_x3*(-d)^3-coef2_x*(-d) - numerator/denominator}}.
\\end{align}
\\]
Esta misma respuesta también se hubiera obtenido si hubiésemos sustituido nuestro valor de $ s $ en la otra ecuación.
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\n
[[0]]$s + t$ = [[1]].
El teorema del resto establece que si un polinomio $f(x)$ se divide por $ (\\simplify{a*x-b})$, entonces el resto es $ f (\\frac {b}{a}) $.
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\n[[0]]$s+t$ = [[1]].
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\n$s =$ [[0]]
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\n$t =$ [[0]]
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\nLuego, reorganiza esta ecuación para encontrar el valor de $t$.
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", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "
The coefficient of friction is $\\mu_s=\\var{mu_s}$ between all surfaces of contact. Knowing that $M_1=\\var{M1}$kg and $M_2=\\var{M2}$kg, determine the smallest force $\\mathbf{P}$ required to start the block $M_2$ moving if:
", "advice": "(a)
\nFree body: $\\var{M1}$-kg block
\n
$W_1=M_1 g=\\var{m1}\\times 9.81=\\var{m1*9.81}\\,\\text{N}$
\n$+\\!\\!\\uparrow\\sum F_y=0:\\quad N_1=W_1$
\n$F_1=\\mu_s N_1=\\var{mu_s}\\times\\var{m1*9.81}=\\var{mu_s*m1*9.81}\\,\\text{N}$
\n$\\overset{+}{\\rightarrow}\\sum F_x=0:\\quad T-F_1=0,\\quad T=F_1=\\var{mu_s*m1*9.81}\\,\\text{N}$
\n\n\nFree body: $\\var{M2}$-kg block
\n
$W_2=M_2g=\\var{M2}\\times 9.81=\\var{m2*9.81}\\,\\text{N}$
\n$+\\!\\!\\uparrow\\sum F_y=0: \\quad N_2-N_1-W_2=0,\\quad N_2=W_1+W_2=\\var{(m1+m2)*9.81}\\,\\text{N}$
\n$F_2=\\mu_s N_2=\\var{mu_s}\\times\\var{(m1+m2)*9.81}=\\var{mu_s*(m1+m2)*9.81}\\,\\text{N}$
\n$\\overset{+}{\\rightarrow}\\sum F_x=0: -P+F_1+F_2=0$
\n\\[P=F_1+F_2=\\var{P1}\\,\\text{N}\\quad\\blacktriangleleft\\]
\n\n(b)
\nBlocks move together (same as in Question 4)
\n\\[P=\\var{P2}\\,\\text{N}\\quad\\blacktriangleleft\\]
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