Processing Math: Done
Solution to Test Paper 1
From Mechanics
(Difference between revisions)
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& \mu =\frac{3 \textrm{.}6}{19 \textrm{.}6}=0 \textrm{.}184 \\ | & \mu =\frac{3 \textrm{.}6}{19 \textrm{.}6}=0 \textrm{.}184 \\ | ||
\end{align}</math> | \end{align}</math> | ||
- | | | + | | (3 marks) |
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| | | | ||
- | | | + | | '''(8 marks)''' |
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| 4 (a) | | 4 (a) | ||
- | | | + | | [[Image:test1ans2.gif]] |
- | | | + | | (1 mark) |
|- | |- | ||
| 4 (b) | | 4 (b) | ||
- | | | + | | <math>\begin{align} |
- | | | + | & R+T\sin 30{}^\circ =200\times 9 \textrm{.}8 \\ |
+ | & R+0 \textrm{.}5T=1960 \\ | ||
+ | & R=1960-0 \textrm{.}5T \\ | ||
+ | \end{align}</math> | ||
+ | | (3 marks) | ||
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| 4 (c) | | 4 (c) | ||
- | | | + | | <math></math> |
- | | | + | <math>\begin{align} |
+ | & F=T\cos 30{}^\circ \\ | ||
+ | & T\cos 30{}^\circ =0\textrm{.}6(1960-0\textrm{.}5T) \\ | ||
+ | & T(\cos 30{}^\circ +0\textrm{.}3)=1176 \\ | ||
+ | & T=\frac{1176}{(\cos 30{}^\circ +0\textrm{.}3)}=1010\text{ N} \\ | ||
+ | \end{align}</math> | ||
+ | | (4 marks) | ||
|- | |- | ||
- | | | + | | |
- | | | + | | |
- | | | + | | '''(8 marks)''' |