Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.1 kgkg gibbon has an arm length (hand to shoulder) of 0.60 mm. We can model its motion as that of a point mass swinging at the end of a 0.60-mm-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.3 m/s. Required:
What upward force must a branch provide to support the swinging gibbon?

254.345 N

Explanation:

Given that

Mass m = 9.1 kg

Velocity, v = 3.3 m/s

Radius, r = 0.6 m

Recall newton's second law of motion,

F = ma, where

F = Force exerted

m = mass of object

a = acceleration of object

Since the forces acting on this body are gravitational and tensile, we then have our F to be

F = T - mg

Also, we have our acceleration to be centripetal and thus,

a = v²/r

Remember that F = ma and F = T - mg. This means that

T - mg = ma.

Substituting the formula for a, we have

T - mg = mv²/r

On solving, we have

T = mv²/r + mg

T = m(v²/r + g)

If we then substitute the values we get from the question into it, we have

T = 9.1 (3.3²/0.6 + 9.8)

T = 9.1 (10.89/0.6 + 9.8)

T = 9.1 (18.15 + 9.8)

T = 9.1 * 27.95

T = 254.345 N

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s