Solution 10.5
From Mechanics
(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>.) |
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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, | ||
+ | |||
+ | <math>\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}</math> | ||
+ | |||
+ | We now use the friction condition <math>F = \mu R</math>, however we first need to calculate <math>R</math>. | ||
+ | |||
+ | 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, | ||
+ | |||
+ | <math>\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}</math> | ||
+ | |||
+ | Using the friction equation gives | ||
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+ | |||
+ | <math>\begin{align} | ||
+ | & 33\textrm{.}3=\mu 171\textrm{.}2 \\ | ||
+ | & \\ | ||
+ | & \mu =\frac{33\textrm{.}3}{171\textrm{.}2}=0\textrm{.}195 \\ | ||
+ | \end{align}</math> |
Current revision
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}