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. what 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.

Explanation:

1) Data:

a) W = 875 N

b) Fx = 300 N (right direction = positive)

c) Fp = ?

d) μs = 0.56

e) μk = 0.47

2) Physical principles and formulae:

a) For sliding, Fp ≥ μs × N, where Fp = μs × N is the magnitud of the minimum force need to exert on the crate to make it start sliding

3) Solution:

a) Free body diagram

The balance of the vertical forces implies that the normal force (N) equals the weight (W) of the crate:

b) Fp = μs × N = 0.56 × 875 N = 490 N ← answer

Remarks:

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

1 is a answer of a question