Question

1.
What is the maximum value of force F such that the block shown in the arrangement, does
not move?
A) 20 N
B) 10 N
C) 12 N
D) 15 N
​

Answer 1

Explanation:

10 N please mark me brainliest

Answer 2

In vertical equilibrium ,

$\sf \Sigma F_y=mg+Fsin\theta\\\\\sf \to N=mg+Fsin\theta$

In horizontal equilibrium ,

$\sf \Sigma F_x=\mu N\\\\\to \sf Fcos\theta = \mu (mg+Fsin\theta)\\\\\to \sf Fcos(60)=\dfrac{1}{2\sqrt{3}} \bigg(\sqrt{3}(10)+Fsin(60)\bigg) \\\\\to \sf \dfrac{F}{2}=5+\dfrac{F}{4}\\\\\to \sf \dfrac{F}{4}=5\\\\\to \sf \orange{F=20\ N}\ \; \bigstar$

So , Maximum force = 20 N