Answer:
1654 kg m/s
Explanation:
The impulse experienced by an object is equal to the product between the force exerted on the object and the time during which the force lasts:
where:
I is the impulse
F is the force exerted on the object
is the time during which the force is applied
For the object in this problem, we have
(force applied)
(time interval)
Therefore, the impulse experienced by the object is:
Answer:
minimal coefficient of static friction:
Explanation:
Once the Tibetan is hanging from the strap, he is exerting a horizontal force on the yak equal to his weight which is the product of his mass times the acceleration of gravity (g) as written below:
The other forces acting on the yak are (see attached diagram):
* the force of gravity on the yak (identified in blue color in the image as ,
* the normal force (indicated in green in the image and identified by the letter "n") of the ledge on the yak as reaction to the yak's weight
* the force of static friction between the yak's hooves and the ledge (pictured in red in the image and identified with )
Since the normal force and the force of gravity on the yak cancel each other (balance - the yak is not moving vertically), the only forces we need to analyse are the force of the Tibetan's weight via the strap, and the force of static friction which should at least be equal in magnitude so the Tibetan doesn't fall. We assume these two forces are acting horizontally (one to the right: the Tibetan's weight, and one to the left: the static friction).
As we said, we want them to be at least equal so thy are in balance.
We recall that the force of static friction is the product of the normal force (n) times the coefficient of static friction (), such that:
In our case these are the forces at play:
So we need to find what is the minimum coefficient of static friction that precludes the Tibetan from falling. We therefore proceed to make an equality between the force of static friction on the yak and the weight of the Tibetan:
and proceed to solve for the coefficient of friction by dividing both sides by "g" (which by the way cancels out), and by the yak's mass:
where we have rounded to four decimal places the periodic number that the quotient generates. Notice that as expected, the coefficient of friction has no units (they all cancelled out in the division).
