Navjyot Kalra
Last Activity: 9 Years ago
(a)
The magnitude of static force of friction is:
Here, m is the mass of the crate, and g is the acceleration due to gravity.
Round off to three significant figures,
fs = 493 N
The magnitude of applied force by worker is F = 412 N , and is smaller than the static force of friction.
Therefore, the crate will not move.
(b)
Assume that force F1 is applied vertically by another worker, the magnitude of static force of friction would be
f’s = μsN’
It is important to note that the normal force in this case would be given as:
Here m is the mass of the crate, and g is the acceleration due to gravity.
The block will start to move when the magnitude of the applied force is equal to the magnitude of static force of friction, that is
Round off to three significant figures,
F1 = 219 N
Therefore, the vertical force applied by the second worker is 219 N .
Therefore, a horizontal force of 81 N would be enough to put the crate in motion.