Find the magnitude of the child's acceleration

[chwala]

Homework Statement

see attached [ Question 6]

Relevant Equations

Mechanics

In my working i have;

For a)

$\tan 55^{\circ} = \dfrac{450}{R}$

$R = \dfrac{450}{\tan 55^{\circ} }= 315 N$

part b) no problem here ...horizontal to left.

c) This is where my real doubt is,
i have using sine rule;

$\dfrac{9.8}{sin 55^{\circ} }= \dfrac{a}{sin 35^{\circ}}$

$a = \dfrac{9.81 \sin 35^{\circ}}{\sin 55^{\circ}} = 6.86$ m/s

There could be an alternative approach.

 

[Orodruin]

Sure, that works, but you could just as well apply the equation for the tangent that you used in (a):
$$
a = g/\tan 55^\circ
$$
Note that
$$
\tan 55^\circ = \sin 55^\circ / \cos 55^\circ
= \sin 55^\circ / \sin(90^\circ - 55^\circ)
\sin 55^\circ / \sin 35^\circ
$$
so the result is obviously the same as what you got

 

[nasu]

Homework Statement: see attached [ Question 6]
Relevant Equations: Mechanics

View attachment 346897


In my working i have;

View attachment 346898

For a)

$\tan 55^{\circ} = \dfrac{450}{R}$

$R = \dfrac{450}{\tan 55^{\circ} }= 315 N$

part b) no problem here ...horizontal to left.

c) This is where my real doubt is,
i have using sine rule;

$\dfrac{9.8}{sin 55^{\circ} }= \dfrac{a}{sin 35^{\circ}}$

$a = \dfrac{9.81 \sin 35^{\circ}}{\sin 55^{\circ}} = 6.86$ m/s

There could be an alternative approach.

The alternative approach may be to use Newton's second law. You know the net force, don't you?

 

[chwala]

The alternative approach may be to use Newton's second law. You know the net force, don't you?

Correct, I have;

$F = ma$
$315 = \dfrac{450}{9.8} a$

$315 = 45.9a$

$a=\dfrac{315}{45.9} = 6.86 m/s$

The confusion on this question was on the value assigned to gravitational force. Its confusing as to whether to use $10 m/s^2$ or $9.81 m/s^2$.

Cheers man!

 

[Lnewqban]

##a=\dfrac{315}{45.9} = 6.86 m/s^2##

Why is the problem stating that the weight of the child acts horizontally?

 

[Orodruin]

Why is the problem stating that the weight of the child acts horizontally?

It ... is not ...

Given that the resultant of P and the child's weight acts horizontally.

My emphasis.

This is necessary to make the problem solvable. Other types of rides, such as a simple back and forth swing, would have a different direction resultant.

 

[Lnewqban]

Thank you, @Orodruin.

 

[chwala]

There's alternative approach to this, will just post directly from phone and amend later, we have to
1. Resolve forces vertically,

$P_y = P \cos 35^0$

$P = 549.2$ N


2. Resolving horizontally,
$P_x = P \sin 35^0$

$P_x = 549.3 \sin 35^0$

$P_x = 315.2$ N