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RekaNice

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## Homework Statement

Example 2.12 : A block pulled on a horizontal table with friction. A block of weight 100 Newtons rests on a horizontal table. The coefficient of static friction between the block and the table is µs = 0.8.

a. If someone pulls on the block with a force F in the direction 37◦ above the horizontal, what is the minimum value of F which will cause the block to slip?

b. If someone pushes on the block with a force F in a direction 37◦ below the horizontal, what is the minimum value of F which will cause the block to slip?

c. Same question as (b), except the direction of the push is 53◦ below the horizontal.

## Homework Equations

Carrying out a similar analysis for part (c), we obtain f = F cos 53◦ = .6F and N = (100 Newtons) + F sin 53◦ = 100 + .8F and thus f N = .6F 100 + .8F (2.7) If we again draw a graph of f /N as a function of F (Fig. 2.21) we see that the limiting value of f /N, as F → ∞, is 0.75. Thus it is clear that the ratio f /N will never attain the value 0.8 no matter how large F is. Therefore it is impossible to cause the block to slip by pushing on it in a direction 53◦ below the horizontal. As we push harder and harder, the normal force magnitude N increases fast enough so that the plane can always exert a large enough frictional force to counterbalance the horizontal component of the applied force F~ . If we set f /N (as given by eqn.( 2.7)) equal to 0.8 and solve for F, we obtain F = −2000 Newtons. Does this negative value of F have any physical significance? The obvious guess is that a negative push should be interpreted as a pull and that we have shown that a pull of 2000 Newtons in a direction 53◦ above the horizontal will just suffice to make the block slip. This is incorrect! If we repeat the analysis of part (a) for the case when the applied force F is a pull in the direction 53◦ above the horizontal, we find that the critical value for slipping is F = 64.5 Newtons. We conclude that the value F = −2000 Newtons has no physical significance. This illustrates the usefulness of drawing graphs like Fig. 2.21 rather than just formally solving equations. Mathematically, the reason why a negative value of F does not correspond to a pull is that when we replace the symbol F by −F the equations describing case (a) do not go over into those describing case (b).

## The Attempt at a Solution

I understand how to solve the problem but what I don't understand here is why a negative pushing force here can't be interpreted as a pull?

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