Question

A block of mass M = 10 kg is sitting on a surface inclined at angle = 45. Given that the coefficient
of static friction is = 0.5 between block and surface, what is the minimum force F necessary to
prevent slipping?​

Answer 1

${\tt{\pink{ANSWER}}}$

35.36N

${\tt{\pink{Explanation}}}$

plz refer to the attachment above

Answer 2

Given,

block of mass, M = 10kg

angle, Ф= 45°

the coefficient of static friction is,  = 0.5

To Find,

the minimum force F necessary to prevent slipping

Solution,

N = mgcos45° = 10 x 10 x $\frac{1}{\sqrt{2} }$ = $\frac{100}{\sqrt{2}}$ N

$f_{s}$ = μ$_{s}$ x N = 0.5 x    $\frac{100}{\sqrt{2}}$ = $\frac{50}{\sqrt{2} }$ N

Driving force =    $\frac{100}{\sqrt{2}}$ N

F$_{min}$ = $\frac{100}{\sqrt{2}}$ $- \frac{50}{\sqrt{2}}$ = 35.36N or $\frac{50}{\sqrt{2}}$ N

∴ Hence, the minimum force F necessary to prevent slipping

is 35.36N or $\frac{50}{\sqrt{2}}$ N.