A man climbs down a rope which is connected to a tree.The breaking stress of the rope is 75℅ of his mass.Find the maximum acceleration he can climb down the rope?
Does the breaking stress here acts as the tension(T)
If so by applying F=ma downwards,
mg-T =ma(a-maximum acceleration)
T=3mg/4
mg/4=ma
a=g/4
Is this correct?
tension at breaking= breaking force
max tension=mass(g-a)
you are correct.
Thank you!
Yes, you are correct. The breaking stress in this context can be assumed to act as the tension (T) in the rope.
To find the maximum acceleration, you can consider the forces acting on the man when he is climbing down the rope.
The gravitational force (mg) is acting downward, and the tension force (T) in the rope is acting upward.
Using Newton's second law (F = ma), you can equate the sum of the forces to the mass (m) multiplied by the maximum acceleration (a).
mg - T = ma
Since the breaking stress is said to be 75% of the man's mass, the tension (T) can be calculated as:
T = (75/100) * m
T = 3m/4
Substituting this value for T back into the equation, we get:
mg - (3m/4) = ma
Now, we can rearrange the equation to solve for the maximum acceleration (a):
mg - (3m/4) = ma
mg = ma + (3m/4)
mg/4 = ma
a = g/4
Therefore, the maximum acceleration the man can have while climbing down the rope is equal to one-fourth (1/4) of the acceleration due to gravity (g).