If bar B were raised slowly, block A would start to slide at the angle = tan -1

Question

If bar B were raised slowly, block A would start to slide at the angle = tan-1(), which was seen in statics to be one way of determining the friction coefficient . Suppose now that the bar is suddenly rotated, starting from the position =0, at constant angular velocity 0. For = 0.5 and r20 = 0.1g, compute the angle at which A slips downward on B1 and compare the result with tan-1 = tan-1(0.5).

(Hint: Set the external forces, as applied to the freebody diagram, equal to the mass of the block times this acceleration.)

Explanation / Answer

now the block has radial acceleration of mrwo2

so along tangential direction net force is zero cause it is rotating with constant angular velocity

so along tangential direction consider force equilibrium

we get N = mgcos

now along radial direction

mgsin - f = mrwo2

mgsin - mgcos = 0.1mg

sin - 0.5cos = 0.1

or 10sin - 5cos = 1

divide by (102 + 52)

cos(+) = -1

so we get + = (2n+1)

so = (2n+1) -

where tan = 2

example for n = 0

= - = 116.565 degrees

and if tan = then = 26.565 degrees

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