Suppose you have to move a heavy crate of weight 875 n by sliding it along a horizontal concrete floor. you push the crate to the right with a horizontal force of magnitude 300 n, but friction prevents the crate from sliding. part adraw a free-body diagram of the crate. part bwhat is the magnitude fp of the minimum force you need to exert on the crate to make it start sliding along the floor? let the coefficient of static friction s between the crate and the floor be 0.56 and that of kinetic friction, k, be 0.47.

The magnitude of the minimum force you need to exert on the crate to make it start sliding along the floor is 490 N.

The given parameters;

The normal force on the crate is calculated as follows;

Fₙ = W = 875 N

The static frictional force on the crate at rest;

Before the crate will move this static frictional force must be overcome.

Thus, the magnitude of the minimum force you need to exert on the crate to make it start sliding along the floor is 490 N.

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A) See figure in attachment

There are 4 forces acting on the crate:

- The horizontal force F, pushing the crate to the right

- The frictional force , acting in the opposite direction (to the left)

- The weight of the crate, , acting downward (with m being the mass of the crate and g the acceleration due to gravity)

- The normal reaction of the floor agains the crate, N, acting upward, and of same magnitude of the weight

The crate is in equilibrium (it is not moving), this means that the forces along the horizontal direction and along the vertical direction are balanced, so:

B) 490 N

In order to make the crate start sliding along the floow, the horizontal push must overcome the maximum force of static friction, which is given by

where

is the coefficient of static friction

N is the normal reaction

The normal reaction is equal to the weight of the crate, so

and so, the maximum force of static friction is

f=ma newton

f=100x2.5=250n

i think the answer is a, hope this : )