Solution 10.5

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(New page: Image:10.5.gif We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector <math>a</math>.)
Current revision (13:54, 23 May 2010) (edit) (undo)
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We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector <math>a</math>.
We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector <math>a</math>.
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Applying Newton´s Second Law <math>F=ma</math> horisontally,
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<math>\begin{align}
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& \to :\ 50\cos {{30}^{\circ }}-F=ma \\
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& \\
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& F=50\cos {{30}^{\circ }}-ma=50\times 0\textrm{.}866-20\times 0.5=43\textrm{.}3-10=33\textrm{.}3\ \text{N} \\
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\end{align}</math>
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We now use the friction condition <math>F = \mu R</math>, however we first need to calculate <math>R</math>.
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Note that <math>R</math> is NOT equal to <math>mg</math> here as the string has a component force in the vertical direction! In fact one has,
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<math>\begin{align}
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& \uparrow :\ R+50\sin {{30}^{\circ }}-mg=0 \\
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& \\
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& R=mg-50\sin {{30}^{\circ }}=20\times 9\textrm{.}81-50\times \frac{1}{2}=196\textrm{.}2-25=171\textrm{.}2\ \text{N} \\
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\end{align}</math>
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Using the friction equation gives
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<math>\begin{align}
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& 33\textrm{.}3=\mu 171\textrm{.}2 \\
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& \\
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& \mu =\frac{33\textrm{.}3}{171\textrm{.}2}=0\textrm{.}195 \\
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\end{align}</math>

Current revision

Image:10.5.gif

We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector \displaystyle a.

Applying Newton´s Second Law \displaystyle F=ma horisontally,

\displaystyle \begin{align} & \to :\ 50\cos {{30}^{\circ }}-F=ma \\ & \\ & F=50\cos {{30}^{\circ }}-ma=50\times 0\textrm{.}866-20\times 0.5=43\textrm{.}3-10=33\textrm{.}3\ \text{N} \\ \end{align}

We now use the friction condition \displaystyle F = \mu R, however we first need to calculate \displaystyle R.

Note that \displaystyle R is NOT equal to \displaystyle mg here as the string has a component force in the vertical direction! In fact one has,

\displaystyle \begin{align} & \uparrow :\ R+50\sin {{30}^{\circ }}-mg=0 \\ & \\ & R=mg-50\sin {{30}^{\circ }}=20\times 9\textrm{.}81-50\times \frac{1}{2}=196\textrm{.}2-25=171\textrm{.}2\ \text{N} \\ \end{align}

Using the friction equation gives


\displaystyle \begin{align} & 33\textrm{.}3=\mu 171\textrm{.}2 \\ & \\ & \mu =\frac{33\textrm{.}3}{171\textrm{.}2}=0\textrm{.}195 \\ \end{align}