Question

The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A

Answer 1

We have that for the Question "" it can be said that Calculate the moment which must be applied to the handle of the screw to raise the block is

• M = 7.30 N.m

From the question we are told

The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A

Generally the equation for the Block is mathematically given as

$\sum Fy=0$

$981cos21.80 = R_2cos53.6\\\\R_2=1535N$

the equation for the Wedge is mathematically given as

$\sum Fx=0\\\\1535cos36.4=Pcos21.8\\\\P=1331N$

the equation for the Screw is mathematically given as

$\beta = tan^{-1}*\frac{L}{2*\pi*r} \\\\\beta = tan^{-1}*\frac{10}{2*\pi*(15)} \\\\\\beta = 6.06\\\\\theta = tan^{-1}*0.25 \\\\\theta = 14.04\\\\\\Therefore\\\\\theta + \beta = 20.1$

Therefore

$M = Pr tan (\theta + \beta)\\\\M = 1331(0.015) tan20.09\\\\M = 7.30 N.m$

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