Finding the force to pull a block up a frictionless slope?

[Eclair_de_XII]

Homework Statement

"A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.50, what minimum force magnitude is required from the rope to start the crate moving? (b) If μ_k = 0.35, what is the magnitude of the initial acceleration of the crate?"

Homework Equations

f_{s, max} = μ_sF_N
f_k = μ_kF_N
$m=68kg$
$θ=255°$
$μ_s=0.5$
$μ_k=0.35$

Answer according to book: (a) 300 N; (b) 1.3 m/s²

The Attempt at a Solution

$F_g=(cos255°)(9.8\frac{m}{s^2})(68kg)=-172.48N$
$F_N=(sin255°)(9.8\frac{m}{s^2})(68kg)=-643.69N$
$f_s=(-643.69N)(0.5)=-321.84N$

But I know this isn't all that is needed to make the block go up. I have to add the difference between the gravitational force and friction, plus the friction, going up, to yield the force to pull the block up. But when I do this, the numbers do not match.

 

[haruspex]

Why does your diagram show friction going both ways?

Edit: as in the other thread, you have created an unknown force called Fg but not defined it. We cannot check whether your equations are right if you do not state what the variables represent.

 

[Eclair_de_XII]

Why does your diagram show friction going both ways?

Because there's friction when the block goes up and goes down.

as in the other thread, you have created an unknown force called Fg but not defined it.

It's gravity, relative to the block's axes.

 

[haruspex]

Because there's friction when the block goes up and goes down.

It will not do both at once. Draw each diagram for a specific context. In this case, I don't see any part of the question where the block will be sliding down.

It's gravity, relative to the block's axes.

Ok, but that's still ambiguous. Do you mean parallel to the slope or normal to it?

 

[Eclair_de_XII]

Do you mean parallel to the slope or normal to it?

It's 255° from the direction of its travel.

 

[Eclair_de_XII]

Never mind; I'm just so bad at this. No one needs to help with this thread anymore.

 

[haruspex]

Never mind; I'm just so bad at this. No one needs to help with this thread anymore.

I just reread the question and realized your diagram is quite wrong.
The floor is level. The rope is at 15 degrees.

 

[Eclair_de_XII]

I'm not doing the problem. I'm too damn tired of this.

 

[Eclair_de_XII]

The floor is level. The rope is at 15 degrees.

Thank you for this insight. Let F = magnitude of the force pulling the block at a 15° angle. I know someone already did this, but it got deleted. Anyway, I want to show my process. This will be the last physics problem I do until the fall. Self-studying just isn't working for me. I get too frustrated and it messes up my mood. It's not good for my health. Thanks, guys, for being so patient with me when I've been in such a pissy mood. Anyway, here's how I did the problem. I expressed $f_s$ in terms of $-(a)(68kg)(cos15°) = F_x$ since the latter opposes the former, and vice versa. Then I express friction as such:

$f_s=(68kg)((a)(sin15°)-9.8\frac{m}{s^2})(0.5)$
$-(a)(68kg)(cos15°)=(34kg)((a)(sin15°)-9.8\frac{m}{s^2})$
$-(a)(68kg)(cos15°)=(34kg)(a)(sin15°)-333.2N$
$-(a)(68kg)(cos15°)-(a)(34kg)(sin15°)=-333.2N$
$-a((68kg)(cos15°)+(34kg)(sin15°))=-333.2N$
$a=\frac{333.2N}{(68kg)(cos15°)+(34kg)(sin15°)}$
$a=4.474\frac{m}{s^2}$
$F=(68kg)(4.474\frac{m}{s^2})=304.2N$
$F_x=(68kg)(cos15°)(4.474\frac{m}{s^2})=293.87N$

$f_k=(68kg)((4.474\frac{m}{s^2})(sin15°)-9.8\frac{m}{s^2})(0.35)=-205.7N$
$F_x-f_k=88.17N$
$a_x=\frac{88.17N}{68kg}=1.297\frac{m}{s^2}$

Again, thank you, everyone. I haven't been acting very grateful recently, I'm sorry to say.

 

[Kaura]

Thank you for this insight. Let F = magnitude of the force pulling the block at a 15° angle. I know someone already did this, but it got deleted. Anyway, I want to show my process. This will be the last physics problem I do until the fall. Self-studying just isn't working for me. I get too frustrated and it messes up my mood. It's not good for my health. Thanks, guys, for being so patient with me when I've been in such a pissy mood. Anyway, here's how I did the problem. I expressed $f_s$ in terms of $-(a)(68kg)(cos15°) = F_x$ since the latter opposes the former, and vice versa. Then I express friction as such:

$f_s=(68kg)((a)(sin15°)-9.8\frac{m}{s^2})(0.5)$
$-(a)(68kg)(cos15°)=(34kg)((a)(sin15°)-9.8\frac{m}{s^2})$
$-(a)(68kg)(cos15°)=(34kg)(a)(sin15°)-333.2N$
$-(a)(68kg)(cos15°)-(a)(34kg)(sin15°)=-333.2N$
$-a((68kg)(cos15°)+(34kg)(sin15°))=-333.2N$
$a=\frac{333.2N}{(68kg)(cos15°)+(34kg)(sin15°)}$
$a=4.474\frac{m}{s^2}$
$F=(68kg)(4.474\frac{m}{s^2})=304.2N$
$F_x=(68kg)(cos15°)(4.474\frac{m}{s^2})=293.87N$

$f_k=(68kg)((4.474\frac{m}{s^2})(sin15°)-9.8\frac{m}{s^2})(0.35)=-205.7N$
$F_x-f_k=88.17N$
$a_x=\frac{88.17N}{68kg}=1.297\frac{m}{s^2}$

Again, thank you, everyone. I haven't been acting very grateful recently, I'm sorry to say.

I worked the problem out earlier and got the same results

Good job and perseverance

 

[haruspex]

Thanks for returning to this and not letting my efforts go to waste.
I found your method a little strange in that you seem to have defined a as tension/mass, not an obvious thing to do. But it works.

 

[Eclair_de_XII]

a defines acceleration of the upward 15° force, and 68 kg defines the mass. That makes a force, which is opposite of friction.

 

[Eclair_de_XII]

Thanks for the tip, by the way; really helped me out.