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    A small fluffy toy hangs on a string in the rear window of a car. The car rounds a bend at 30ms-1 and the string hangs 20 degrees from the vertical.

    There is a diagram just of the string hanging 20 degrees from vertical.


    b) Calculate the radius of the curve in the road.
    sin 20=mv^2/r
    cos 20=mg
    i dont know where the sin and cos came otherwise i know how to do this question
    sry i always get mixed up when to use cos and sin
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    (Original post by SOREN_TORETTO)
    A small fluffy toy hangs on a string in the rear window of a car. The car rounds a bend at 30ms-1 and the string hangs 20 degrees from the vertical.

    There is a diagram just of the string hanging 20 degrees from vertical.


    b) Calculate the radius of the curve in the road.
    sin 20=mv^2/r
    cos 20=mg
    i dont know where the sin and cos came otherwise i know how to do this question
    sry i always get mixed up when to use cos and sin
    Have you drawn a force diagram? I'm assuming you need to calculate the forces perpendicular to the string and make the centripetal and gravitational components equal.
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    (Original post by Chlorophile)
    Have you drawn a force diagram? I'm assuming you need to calculate the forces perpendicular to the string and make the centripetal and gravitational components equal.
    yeah tension upwards and mg down
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    (Original post by Chlorophile)
    Have you drawn a force diagram? I'm assuming you need to calculate the forces perpendicular to the string and make the centripetal and gravitational components equal.
    i dont get what to use whether sin or cos
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    (Original post by SOREN_TORETTO)
    yeah tension upwards and mg down
    Well you've also got the centripetal force to the left. The tension isn't doing anything, it's parallel to the line of movement.
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    The tension in the string (it's horizontal component) is providing the centripetal force.
    The vertical component of the tension balances the weight of the toy.

    It's these two components that the sin and cos come from.

    You should have Tsin theta and Tcos theta in those equations.
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    (Original post by Stonebridge)
    The tension in the string (it's horizontal component) is providing the centripetal force.
    The vertical component of the tension balances the weight of the toy.

    It's these two components that the sin and cos come from.

    You should have Tsin theta and Tcos theta in those equations.
    Ty i got it but im confused still in when to use sin or cos
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    (Original post by SOREN_TORETTO)
    Ty i got it but im confused still in when to use sin or cos
    Well cos is adjacent divided by hypotenuse, and the component adjacent to the angle given is always cos.

    The question says the spring hangs at 20 degs to the vertical.
    This means the 20 deg angle of the string is adjacent to the vertical direction.
    So the vertical component will be cos in this question.
    That is why the equation shows the vertical component of the tension is T cos theta and this balances the weight, which is in the vertical direction.
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    Thx alot now i get it
 
 
 
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