Question

A person is pushing against a wall while their body is slanted. The wall is made of smooth ice, therefore it has zero friction. The floor, however, is made of rough sandstone and as a result, has coefficients of friction $\mu _{s}$ and $\mu _{k}$ , both of which are non-zero.

Model the person as a uniform one-dimensional rod of mass m and length L. Their body makes an angle $\theta$ relative to the ground as shown.

Known: m, L, $\theta$ ,   $\mu _{s}$ , $\mu _{k}$ , g

Note: There is no friction between the wall and the person.

Part 1: If the person is stationary without slipping, determine the magnitude of the normal force exerted on the person by the wall.

Part 2: Now the person tries to lean more horizontal so that their feet get closer and closer to slipping against the floor. What is the smallest angle $\theta _{smallest}$ relative to the ground that they could lean without slipping?

Answer 1

Partal FBD Nw rotational) point As person is in equilibrium (both translational and so het Torque about any should be zero. Calculating Net Torque about B :- a) mg I coso – No Lring=0 - No = mgh coso L sine » Nw = ng coto Part 2 for smallest angle smallent) relative to ground so that person could lean without slipping, friction acting must be maximum static friction. For vertical equilibrium! Na-ing = 0 Na = ng Maximum Static friction & Asna = Ms ng

still . person will be in Net Torque, about rotational equilibrium as well. I should also be zero. EZA = 0 =) - ing & coso - flsino & NG Lcoso =0 Substituting, f= Msing and Nang = -mg cose - Ms uptsino & my Low=0 - coso - ussino & coro = 0 - , ius - coto= 2ms - o- cot (2 ps)