Question #272477

A 5.0m long ladder who weighs 180N rests against a vertical wall, making an angle of 53 degrees with the ground. When a person who weighs 800N climbs this uniform ladder and stops one-third of the way up the ladder. Find the normal and frictional forces on the ladder at its base.


1
Expert's answer
2021-11-29T11:44:06-0500

Fy=0: NA=980NA=980 (N)\sum F_y=0:\ N_A=980\to N_A=980\ (N) . Answer


Fx=0: NB+Ffr A=0Ffr A=NB\sum F_x=0:\ -N_B+F_{fr\ A}=0\to F_{fr\ A}=N_B


MA=0: 1801.5+80024NB=0NB=268 (N)\sum M_A=0:\ 180\cdot1.5+800\cdot2-4N_B=0\to N_B=268\ (N)


Ffr A=NB=268 (N)F_{fr\ A}=N_B=268\ (N) . Answer






Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS