Find Force of Friction on Wedge: Shm Physics Help

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SUMMARY

The discussion focuses on calculating the force of friction between a wedge and the ground in a physics problem involving a mass m attached to a spring on a wedge inclined at 60 degrees. The equations of motion derived include F=-kx and m(d²x/dt²)=-1/2 kx, leading to the conclusion that the force of friction is f_r=maω²sin(ωt)/2. A clarification is made regarding the use of 'a' versus 'A', where 'A' refers to amplitude. The final answer aligns with the book's solution, confirming the correctness of the calculations.

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Homework Statement


A wedge of mass M is placed on a rough surface with a mass m attached to a spring placed rigidly on the wedge. The surface b/w the wedge and the mass m is smooth and the surface between the wegde and the ground is sufficiently rough to keep the wedge from moving. The angle of inclination of the wedge w.r.t ground is 60 degrees. The small block is displaced by a small distance x. Find the force of friction between the ground and the wedge.


Homework Equations



[tex]\frac{d^2 x}{dt^2)=-kx[/tex]
[tex]x=asin(\omega t)[/tex]

The Attempt at a Solution


Force acting on the mass m :

[tex]F=-kx[/tex]
Force along the horizontal direction:
[tex]F=-kxcos(60)[/tex]
[tex]F=-\frac{1}{2}kx[/tex]
[tex]m\frac{d^2x}{dt^2}=-\frac{1}{2}kx[/tex]
[tex]\frac{d^2x}{dt^2}=-\frac{1 k}{2 m}x[/tex]
[tex]a=-\frac{k}{2m}asin(\omega t)[/tex]
[tex]\frac{k}{m}=\omega ^2[/tex]
[tex]f_r=Ma[/tex]
[tex]f_r=\frac{1 Ma\omega ^2sin(\omega t)}{2}[/tex]

but the answer given is [tex]f_r=\frac{ma\omega ^2sin(\omega t)}{2}[/tex]
Is the book wrong or have I made a mistake?
 
Last edited:
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Are you using a to mean two different things here?
 
Yeah, sorry. In the end it should be A, for amplitude.
 

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