Question #09151

1 Answer
Cesareo R.
Dec 11, 2016

#M=34.9677# [kg]

Explanation:

Calling #T# the tension on the rope, #N_1# the normal force at the man floor, #N_2# the normal force at the mass location, #mu_s# the friction coefficient, #m# the man mass #theta# the angle at the rope and #g# the gravity, we have

Horizontally

#T cos(theta)-mu_s N_1=0#

Vertically

#N_1-m g + T sin(theta)=0# and #T+N_2-M g=0#

but at the problem conditions we have #N_2=0# so we have

#{(T cos(theta)-mu_s N_1=0),(N_1-m g + T sin(theta)=0),(T-M g=0):}#

Solving for #T,N_1,M# we obtain

#((T = (g m mu_s)/(Cos(theta) + mu_s Sin(theta))),(N_1 = (g m Cos(theta))/(Cos(theta) + mu_s Sin(theta))),(M = (m mu_s)/(Cos(theta) + mu_s Sin(theta))))#

So finally

#M = (m mu_s)/(Cos(theta) + mu_s Sin(theta))=34.9677#

Related questions
See all questions in Defining Force
Impact of this question
1350 views around the world
You can reuse this answer
Creative Commons License