Figure given along side shows a block pressed against a rough vertical wall with a force F as shown in side view . Column -I shows angle at which force F is applied and Column-II givesinformation about corresponding friction force , match them .

Figure given along side shows a block pressed against a rough vertical

1.	Figure given along side shows a block pressed against a rough vertical wall with a force F as shown in side view . Column -I shows angle at which force F is applied and Column-II givesinformation about corresponding friction force , match them .
Answer» Solution :If `F sin theta lt mg `, f is upward . `sin theta lt (75)/(100) , sin theta lt (3)/(4) , thetalt 53^(@)` (c) `theta = 53^(@) , F sin 53^(@) = 100 XX (4)/(5) =80 N` `therefore f = 80 - 75 = 5 N` (STATIC, downward) `therefore(c) to (q) , (r)` (d) `theta = 0^(@) , F sin 0^(@) = 0 ( therefore f` is upward) `f_("limit") = mu N= mu F cos theta = 0.1 xx 100 xx (4)/(5) = 8 N` Thus, friction is kinetic because (0 + 8) N is less than 75 N , `therefore (d) to (p) , (s)` (a) `theta = 37^(@) , F sin theta = 100 xx (3)/(5) = 60 N( therefore` f is upward) mg = 75 N `f_("limit") = 8 N` `therefore ` f is kinetic and upward. (b) `theta = 45^(@) , F = sin 45^(@) = (100)/(SQRT2) = 50 sqrt2 = 70.7 N` `f_("limit") = 8 N` Thus , `mg lt (F sin theta + f_("limit") )` and f is static and upward .