گاما رو نصب کن!

{{ number }}
اعلان ها
اعلان جدیدی وجود ندارد!
کاربر جدید

جستجو

پربازدیدها: #{{ tag.title }}

جستجوهای پرتکرار

میتونی لایو بذاری!

Particles $P$ and $Q$ have masses $0.8{\text{ }}kg$ and $0.4{\text{ }}kg$ respectively. $P$ is attached to a fixed point $A$ by a light inextensible string which is inclined at an angle ${\alpha ^ \circ }$ to the vertical. $Q$ is attached to a fixed point $B$, which is vertically below $A4, by a light inextensible string of length $0.3{\text{ }}m$. The string $BQ$ is horizontal. $P$ and $Q$ are joined to each other by a light inextensible string which is vertical. The particles rotate in horizontal circles of radius $0.3{\text{ }}m$ about the axis through $A$ and $B$ with constant angular speed $5{\text{ }}rad{\text{ }}{s^{ - 1}}$ (see diagram).

a) By considering the motion of $Q$, find the tensions in the strings $PQ$ and $BQ$.

b) Find the tension in the string $AP$ and the value of $\alpha $.

پاسخ تشریحی :
نمایش پاسخ

a) ${T_{PQ}} = \left( {0.4g} \right) = 4N$

${T_{BQ}} = 0.4 \times {5^2} \times 0.3$

${T_{BQ}} = 3N$

b) $T\cos \alpha  = 0.8g + 4$

$T\sin \alpha  = 0.8 \times {5^2} \times 0.3$

${T^2} = {12^2} + {6^2}$

${T_{AP}} = 13.4N{\text{ }}\left( { = 6\sqrt 5 N} \right)$

${\alpha ^ \circ } = {\tan ^{ - 1}}{\text{ }}\left( {6/12} \right) = {\tan ^{ - 1}}\left( {1/2} \right) = {26.6^ \circ }$

OR $T\cos \alpha  = 0.8g + 4$

$T\sin \alpha  = 0.8x{5^2} \times 0.3$

$\tan \alpha  = 6/12$

$\alpha  = 26.6$

${T_{AP}} = 13.4N$

تحلیل ویدئویی تست

تحلیل ویدئویی برای این تست ثبت نشده است!