Answer:
[tex]F_a = F_b = F_c = F_d = F_e = F_f[/tex]
all will have same frictional force on them
Explanation:
As we know that the crate is placed at rest on the rough floor
so here the forces must be balanced
So here the applied force must be equal to the friction force on the crate
So we will say that the friction force on the crate is static friction and it must be equal to the applied force on the crate
So all the crates have same external force which means that all the crates will have same frictional force on them
so we have
[tex]F_a = F_b = F_c = F_d = F_e = F_f[/tex]
all will have same frictional force on them
This question involves the concepts of frictional force and coefficient of static friction.
The ranking of the frictional force acting on the crates will be[tex]"f_d>f_e=f_f>f_a=f_b>f_c"[/tex].
The frictional force can be calculated by the following simple formula:
[tex]f = \mu R[/tex]
where,
f = frictional force = ?
[tex]\mu[/tex] = coefficient of static friction (since boxes are not moving)
R = Normal Reaction = Weight of Box = mg
Now, we use this formula to calculate the frictional force acting on each box.
BOX A:
[tex]f_a=\mu mg = (500\ kg)(0.6)g = 300\ g[/tex]
BOX B:
[tex]f_b=\mu mg = (750\ kg)(0.4)g = 300\ g[/tex]
BOX C:
[tex]f_c=\mu mg = (250\ kg)(0.2)g = 50\ g[/tex]
BOX D:
[tex]f_d=\mu mg = (600\ kg)(0.8)g = 480\ g[/tex]
BOX E:
[tex]f_e=\mu mg = (750\ kg)(0.6)g = 450\ g[/tex]
BOX F:
[tex]f_f=\mu mg = (1500\ kg)(0.3)g = 450\ g[/tex]
Therefore, the ranking will be:
[tex]"f_d>f_e=f_f>f_a=f_b>f_c"[/tex]
Learn more about friction force here:
https://brainly.com/question/1714663?referrer=searchResults
The attached picture shows the friction force.
