Question

A smooth right anyoed rigid triangular frame ABC made of thin rod is fixed in a vertical plane​

Answer 1

Answer:

A small bead starting from rest from corner $A$ takes equal time t to slide down the arms $A B$ and $A C$. If $B C=1$ then $\frac{2 l}{g}$.

Explanation:

A smooth right angled rigid triangular frame ABC made of thin rod is fixed in a vertical plane​. A rigid triangular frame $\mathrm{ABC}$ of mass $\mathrm{m}$ is hanging from a rigid horizontal rod AB. The frame is constrained to move along horizontal without friction. A bead of mass $\mathrm{m}$ is released from B that moves along BC. A small bead starting from rest from corner A takes equal time t to slide down the arms $A B$ and $A C$. If $B C=1$.

$\begin{aligned}& t=\frac{2 A B}{g \sin \left(\phi_1-16^{\circ}\right)} \\& t=\frac{2 A C}{g \sin \left(\phi_2+16^{\circ}\right)} \\& \frac{A B}{\sin \left(\phi_1-16^{\circ}\right)}=\frac{A C}{\sin \left(\phi_2-16^{\circ}\right)} \\& \frac{\sin d_2}{\sin \left(\phi_1-16^{\circ}\right)}=\frac{\sin d_1}{\sin \left(\phi_2+16^{\circ}\right)} \\& \phi_1=53^{\circ}, \phi_2=37^{\circ} \\& t=\frac{2 l \times 3}{5 \times g \sin \left(37^{\circ}\right)}=\frac{2 l}{g}\end{aligned}$

Therefore, A small bead starting from rest from corner $A$ takes equal time t to slide down the arms $A B$ and $A C$. If $B C=1$ then $\frac{2 l}{g}$.

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