What is the smallest vertical force which will move the crate

[Air]

Homework Statement

A crate with a weight of $50N$ rests on a horizontal surface. A person pulls horizontally on it with a force of $10N$ and it does not move. To start it moving a second person pulls vertically upwards on the crate. If the coefficient of static friction is $0.4$, what is the smallest vertical force which will move the crate.

Homework Equations

$f_s=\mu_sn$

The Attempt at a Solution

$R(\rightarrow): 10 - f_s = 0 \implies f_s = 10N$
$f_s=\mu_sn_{tot} = (0.4)(n_{tot}) = 10 \implies n_{tot} = 25$
$n_{tot} = F_y+n \implies F_y = 25-50 =-25N$

4. The problem I encounter
Can you check my method please. I got $-25N$ which is negative hence this is giving me a doubt. Thanks.

 

[Stovebolt]

Homework Statement

A crate with a weight of $50N$ rests on a horizontal surface. A person pulls horizontally on it with a force of $10N$ and it does not move. To start it moving a second person pulls vertically upwards on the crate. If the coefficient of static friction is $0.4$, what is the smallest vertical force which will move the crate.

Homework Equations

$f_s=\mu_sn$

The Attempt at a Solution

$R(\rightarrow): 10 - f_s = 0 \implies f_s = 10N$
$f_s=\mu_sn_{tot} = (0.4)(n_{tot}) = 10 \implies n_{tot} = 25$
$n_{tot} = F_y+n \implies F_y = 25-50 =-25N$

4. The problem I encounter
Can you check my method please. I got $-25N$ which is negative hence this is giving me a doubt. Thanks.

You did everything correctly. The reason for the negative is that your calculations used a downward direction as positive. Thus, your -25 N answer means the 25 N force will be applied upwards.

 

[PhanthomJay]

What is $$n_{tot}$$? The normal force is just $$n$$ . The applied upward force is $$F_y$$. The crates weight is 50N down. You've got a couple of 'n's' in your equation that need to be corrected.

 

[Air]

You did everything correctly. The reason for the negative is that your calculations used a downward direction as positive. Thus, your -25 N answer means the 25 N force will be applied upwards.

How can you tell that I took downwards as positive? Even if I resolve upwards I would get: $R(\uparrow) : n - 50 = 0 \implies n=50N$?

What is $$n_{tot}$$? The normal force is just $$n$$ . The applied upward force is $$F_y$$. The crates weight is 50N down. You've got a couple of 'n's' in your equation that need to be corrected.

My $n_{tot}$ is just a notation I gave for the total force in the upper direction which would be the $F_y + n$. I should have given that another notation but what's wrong other than that, is my calculation correct?

 

[Doc Al]

$R(\rightarrow): 10 - f_s = 0 \implies f_s = 10N$
$f_s=\mu_sn_{tot} = (0.4)(n_{tot}) = 10 \implies n_{tot} = 25$

The static friction is μn, where n is the normal force (not n_tot). So μn = 10, thus n = 25.

$$n_{tot} = F_y+n \implies F_y = 25-50 =-25N$$

The vertical forces must add to zero:
Fy + n - 50 = 0 (Where Fy is the applied vertical force.)
Solve for Fy.

 

[Air]

The static friction is μn, where n is the normal force (not n_tot). So μn = 10, thus n = 25.

The vertical forces must add to zero:
Fy + n - 50 = 0 (Where Fy is the applied vertical force.)
Solve for Fy.

Ah yes, that gives a positive answer, $F_y = 50 - 25 = 25N$.

The problem was that I took the vertical force combined with the normal force to equal the normal force and used that in the equation.

It makes sence. Thanks to everyone for the help.