A 3.00-m-long, 190-N, uniform rod at the zoo is held in a horizontal position by

Question

A 3.00-m-long, 190-N, uniform rod at the zoo is held in a horizontal position by two ropes at its ends in (Figure 1). The left rope makes an angle of 150 degree with the rod and the right rope makes an angle theta with the horizontal. A 90-N howler monkey (Alouatta seniculus) hangs motionless 0.50 m from the right end of the rod as he carefully studies you. Make a free-body diagram of the rod. Draw the vectors starting at the black dots. The location and orientation of be graded. The length of the vectors will not be graded. Calculate the tension in the left rope. Calculate the tension in the right rope.

Explanation / Answer

current, i1=5A

current, i2=2A

separation,r=0.4 m

length of the rod, l=3m

weight of the rod, w1=190 N

weight of the monkey, w2=90 N

(at x=0.5m from right end)

a)

with respect right end, torque equilibrium,

l*T1*sin(30)=w1*l/2+w2*0.5

3*T1*sin(30)=190*1.5+90*0.5

==> T1=220 N

b)

with respect left end, torque equilibrium,

l*T2*sin(theta)=w1*l/2+w2*2.5

3*T2*sin(theta)=190*1.5+90*2.5

====> 3*T2*sin(theta)=510

T2*sin(theta)=510/3

====> T2*sin(theta)=170 N -----(1)

and

T1*cos(30)=T2*cos(theta)

220*cos(30)=T2*cos(theta)

===> T2*cos(thet2)=190.5 N -----(2)

from (1) and (2)

T2^2=170^2+190.5^2

==> T2=sqrt(170^2+190.5^2)

===> T2=255.3 N

and

T2*cos(theta)=190.5

255.3*cos(theta)=190.5

===> theta=41.74 degrees,

and

tension, T2=255.3 N

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