Respuesta :
Answer:
Part a)
[tex]\theta = tan^{-1}\frac{a}{g}[/tex]
so here the angle made by the string is independent of the mass
Part b)
[tex]a = 4.16 m/s^2[/tex]
Explanation:
Part a)
Let the string makes some angle with the vertical so we have force equation given as
[tex]Tcos\theta = mg[/tex]
[tex]T sin\theta = ma[/tex]
so we will have
[tex]tan\theta = \frac{ma}{mg}[/tex]
[tex]\theta = tan^{-1}\frac{a}{g}[/tex]
so here the angle made by the string is independent of the mass
Part b)
Now from above equation if we know that angle made by the string is
[tex]\theta = 23 degree[/tex]
so we will have
[tex]tan23 = \frac{a}{g}[/tex]
[tex]a = g tan23[/tex]
[tex]a = 9.81(tan23)[/tex]
[tex]a = 4.16 m/s^2[/tex]
Answer:
(a) g tan u
(b) 4.16 m/s^2
Explanation:
(a) Let the acceleration of the car is a.
Due to the psheudo force, the mass moves back.
According to the diagram
tan u = ma / mg
tan u = a / g
a = g tan u
(b) u = 23°
a = g tan 23°
a = 9.8 x tan 23°
a = 4.16 m/s^2