Stuck on part b?

enter image source here

I found the angle between the floor and the beam to be about #34.69# degrees and the angle between the beam and weight to be about #55.31# degrees. Lastly, the magnitude of #vecP# to be #125.81N#

1 Answer
Nov 17, 2017

enter image source here

In the above figure forces acting on the beam at different points at a given moment of its lift have been shown.

Here #sintheta =(XY)/(OY)=1.4/2.46#

and #costheta=sqrt(1-sin^2theta)~~2.023/2.46#

Resolved parts of the applied force #P# are #P*costheta and P*sintheta# in vertical and horizontal directions respectively.

Considering considering the moments of the forces about point #O# at equilibrium we can write

#PxxOY=WxxOCcostheta#

#=>Pxx2.46=306xx1.23xx2.023/2.46#

#=>P=306xx2.023/(2xx2.46)~~125.82N#

Now considering the equilibrium of vertical forces we get the reactionary force # R# exerted by floor on the beam

#R+Pcostheta=W#
#=>R=W-Pcostheta#

#=>R=306-125.82*2.023/2.46#
#=>R~~202.53N#

Again considering the equilibrium of horizontal forces we get the frictional force (#F#) acting on the beam to resist its slipping at the given instant.

#F=P*sintheta=125.82*1.4/2.46#
#=>F=71.60N#