What is the maximum tension in the makeshift rope for a jail escape attempt?

[777lov3r]

A 75 kg petty third wants to escape from a third story jail window. Unfortunately, a makeshift rope of sheets tied together can support a mass of only 50 kg. What is the minimal acceleration with which the thief must lower himself so he may use the "rope" without breaking it?

the variables i got were:
m=75 kg
a= -9.80m/s^2
v1=0

i then tried to use the formula for the thief, f=ma, isolated for a, a=f/m and substitued it into the rops max acceraltion limit, which formula is f=ma

i then got f(thief)/m(thief)= f(rope)/m, and i don't really know where to continue from there, because the thief must accelerate at a slower rate than what gravity plans for him, and must be a lower acceleration than 50 kg would normally fall right? please help!

 

[PhanthomJay]

You must draw free body diagrams to designate all forces acting on the person, then apply Newton's laws. Newton 2 is not F=ma, its F_net =ma , or, alternatively, sum of all forces = ma. What forces act on the man ? What's the net force acting on him? Then use Newton 2 to find his required minimum acceleration.

 

[777lov3r]

but I'm confused on whether or not to use that mans mass, 75 kg, if I draw a free body diagram then fnet=fg-fn, but fnet =ma, so I isolate for a? my teacher doesn't really help me out very much..

 

[PhanthomJay]

but I'm confused on whether or not to use that mans mass, 75 kg, if I draw a free body diagram then fnet=fg-fn, but fnet =ma, so I isolate for a? my teacher doesn't really help me out very much..

Yes, you must use the man's mass and weight in the equation. Your equation is correct, (although what you call fn is the allowable tension force in the cable, you should just call it 'T').. Once you know 'fg', solve for 'a' by isolating it.

 

[777lov3r]

oooohh, so we have to find his minimal acceleration , and the tension on the rope in order for it not to break , so it would be a=(fg-ft)/m?but isn't tension unkown?

 

[PhanthomJay]

It is known, indirectly. It is stated that the sheet can only hold a mass of 50 kg (at rest). So what is the maximum tension in the cable just before it breaks?