An Adventurous archaeologist crosses between two rock cliffs byslowly going betw

Question

An Adventurous archaeologist crosses between two rock cliffs byslowly going between the cliffs. He stops to rest at themiddle of the rope (figure 5.38). The rope will break if thetension in it exceeds 2.50x10^4 N. Our hero's mass is 90.0kg. (a) If the angle is 10.0 degrees, find the tensionin the rope. Start with a free body diagram of thearchaeologist. (b) What is the smallest value the angle can have if the rope is not to break?

So I found the first answer to this question by drawing a FBD andsetting up the following equation.

fy=(m)(ay)

Since the person is at rest, there is no acceleration, sotherefore,

fy=0

Since there were 2 forces in the FBD, I came up with thefollowing:

T(2Cos10)-882N=0

T(2Cos10)=882N

To isolate T,

T=882N/2Cos10

T=2.54x10^3 N

I got stuck when I had to find the smallest angle where the ropewould not break. I have a rough idea of how to set it up, butI need some guidance. The critical angle is 1.01 degrees,because i looked it up at the end of the chapter. Pleasehelp!

Explanation / Answer

Given : Mass (m ) = 90 kg Tention ( T ) = 2.5 * 104 N g = 9.8 m/s2 The vertical component     2T sin = m g     = sin-1 [ m g / 2T ]         = ------ Solve it I hope it helps you

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