91Ó°ÊÓ

• $\text{¸é²¹»å¾±³Ü²õ=9.1″¾}$
  
• $\text{³Õ±ð±ô´Ç³¦¾±³Ù²â=16 k³¾/³ó°ù}$
  

The problem deals with Newton’s laws of motion which describe the relations between the forces acting on a body and the motion of the body.

Formula:

$\left|\mathrm{a}\right|={\mathrm{v}}^2/\mathrm{R}$

The free-body diagram (for the hand straps of mass m) is the view that a passenger might see if she was looking forward and the streetcar was curving towards the right (so it points rightwards in the figure).

We note that $\left|\mathrm{a}\right|={\mathrm{v}}^2/\mathrm{R}$

Where,

$\begin{array}{c}\mathrm{v}=\text{16 k³¾/³ó°ù}\\ =4.4\text{″¾/²õ}\end{array}$

Applying Newton’s law to the axes of the problem (+x is rightward and +y is upward), we obtain,

$\mathrm{T}\text{sin}\mathrm{θ}={\mathrm{mv}}^2/\mathrm{R}$

$\mathrm{T}\mathrm{³¦´Ç²õθ}=\mathrm{mg}$

Solving these equations for the angle,

$\mathrm{θ}={\text{tan}}^{−1}\left(\frac{{\mathrm{v}}^2}{\mathrm{Rg}}\right)$

Substitute the values in the above equation, and we get,

$\begin{array}{c}\mathrm{θ}={\text{tan}}^{−1}\left(\frac{{\left(4.4\text{″¾/²õ}\right)}^2}{\left(\text{9.1″¾}\right)\left(9.8{\text{″¾/²õ}}^2\right)}\right)\\ \mathrm{θ}={\text{tan}}^{−1}\left(\frac{{\left(4.4\text{″¾/²õ}\right)}^2}{\left(\text{9.1″¾}\right)\left(9.8{\text{″¾/²õ}}^2\right)}\right)\\ \mathrm{θ}={\text{tan}}^{−1}\left(0.2170\right)\\ \mathrm{θ}=12°\end{array}$

Thus,the vertical angle will be made by the loosely hanging hand straps is $12°$.