Answer to Question #130106 in Mechanics | Relativity for Ariyo Emmanuel

Question

A 50-cm × 30-cm × 20-cm block weighing 150 N is to be moved at a
constant velocity of 0.80 m/s on an inclined surface with a friction
coefficient of 0.27. (a) Determine the force F that needs to be applied
in the horizontal direction. (b) If a 0.40-mm-thick oil film with a
dynamic viscosity of 0.012 Pa∙s is applied between the block and
inclined surface, determine the percent reduction in the required
force.

Answer 1

$\sum f_x=0 and \sum f_y=0$. For direction Y:

$F_n \cos20\degree-F_r\sin 20\degree-W=0.$

$F_n (\cos20\degree-\mu R\sin 20\degree)-W=0$

$F_n(0.9396-0.27\times0.342)-150=0$ Thus $F_n=177.02N$

For direction X:

$F_1=177.02(0.3420+0.27\times0.9396)=105.44N$.

Due to oil application on the block, the shear force on an inclined surface replaces shear force. Thus:

For direction Y:

$\sum F_y=0$. Thus, $F_n\cos20\degree-0.012\times0.5\times0.2 (\frac{0.8}{4\times10^{-4}}\times sin20\degree)-150=0$ Therefore, $F_N=160.5N$ .

For direction X:

$F_2-160.5\sin20\degree-0.012\times0.5\times0.2 (\frac {0.8}{4\times10^4}\cos20)$ .

$F_2=57.2N$

Reduction=$\frac {F_1-F_2}{F_1}=\frac {105.5-57.2} {105}=45$

Thus; Reduction= 45.79%