Statics equilibrium?

3 Answers
Mar 24, 2017

Answer:

#Mg#

Explanation:

Calling

#f_s=(sqrt2/2,sqrt2/2)t_s#
#f_h=(-1,0)t_h#
#p=(0,-Mg)#

the equilibrium equation is

#f_s+f_h+p=(0,0)#

or

#{(sqrt2/2 t_s-t_h=0),(sqrt2/2 t_s-Mg=0):}#

solving for #t_s,t_h# we obtain

#t_s = sqrt2 M g# and
#t_h=Mg#

Mar 25, 2017

Answer:

#F=Mg#

Explanation:

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#tan beta=(Mg)/F#

#F=(Mg)/tan beta#

#tan 45^o=1#

#F=(Mg)/1#

#F=Mg#

Mar 29, 2017

Answer:

Option (1)

Explanation:

my comp
Let #AB# be initial position of mass #M# and #AB'# be its final position.

We see that the mass has moved a horizontal distance #=CB'#
and consequently it has moved through a vertical distance of #BC#
Due to movement in the vertical direction it has gained
Gravitational potential energy#=Mg|vec(BC)|# ......(1)
Due to Law of conservation of energy this must be equal to work done by an effective Horizontal force #vecF# through a distance #vec(CB')#

I have used effective as position of the force keeps on changing as the mass moves up.
Work done by the force#=vecFcdotvec(CB') #

As shown in the fig above, angle between the two is #0^@#
#=># Work done by the force#=|vecF||vec(CB')| # .......(2)

Equating (1) and (2) we get
#Mgbar(BC)=|vecF|bar(CB') #
#=>|vecF| =(Mgbar(BC))/ bar(CB') # ......(3)

Let #L# be length of the string. In #Delta ACB'#
#bar(CB')=L sin45^@=L/sqrt2#
and #bar(BC)=L-L/sqrt2# (#Delta ACB'# is an isosceles triangle)

Inserting these values in (3) we get
#|vecF| =(Mg) (L-L/sqrt2)/ (L/sqrt2) #
#=>|vecF| =Mg (sqrt2-1) #