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Solution Example 1: The object on the table would require a minimum force of $Mgk$. Example 2: The simplest way to the able the amount of force required to move the door hinge that has friction would be to look at a cross section of the component. The cross section would be a circle with two forces being applied. These force would cause torque which will be ...


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Take a look at the illustration below, which is how I understand your question. You're asking, what would be the minimum value for the blue arrows in order to make each object move? First, let's look at the simpler case - the red box. I don't think this needs another diagram - you've rightly stated in your question that if there is a coefficient of friction ...


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The minimum force required to swing the door open would not be equal to the minimum force to push the object off the table. Pushing the door that is attached to the door hinge would create different minimums of force required to push the door open as you could push the door at many different areas. In order to push a door open, you would need a certain ...


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You know the velocity $v$ of point $A$. The angular velocity $\omega$ can (in planar kinematics) be derived from the distance of the point to its center of rotation and its velocity (generally $\vec{v}_A = \vec{\omega} \times \vec{r}_A$ or $v_A=\omega \cdot r_A$ in 2D). The center of rotation of a non-sliding circle is always its contact point $P$ with the ...


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v = sqrt(2) x w, because the disc is instantaneously rotating about the point of contact with the ground and A is at a distance of sqrt(2)XR (from Pythagoras theorem), hence w and v have that relation. Mind that v is the velocity of A (as in diagram) and not the centre of mass.


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