# Newton's 3 Laws Problem

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1. Oct 7, 2014

### paytona

1. The problem statement, all variables and given/known data
The 20.0 kg box is being pushed with a 45.0N Force acting at an angle of 25.0*. The coefficient of kinetic friction on this surface is 0.150. Find the acceleration

2. Relevant equations
F=ma
Fy=0
Fx=0

3. The attempt at a solution

So I attempted this so far

(sum)Fy=0
Py+Fn-Fg=0
Fn=Fg-Py
Fn=(20.0kgx9.81m/s^2)-45.0Nsin335*
Fn=215.217822N

I tried to go and do sum of all the forces on the x axis but there's no point because we already have Ffk and Px.......... So now do I go ahead and put all of the forces in for Sum of F in
(sum)F=ma and solve for a?
as in would I do
Fx-Fg+Fn-Ffk=ma???

2. Oct 7, 2014

### Bedeirnur

Well, actually, as my physics professor always says, do not put in the numbers, if not when you've arrived to the last formula.

So...The box is moving on the x axis, so we know that the sum of forces on the y axis is 0.

We have |Fp| = |N| +|Fy| With fp being the force exerted by gravity on the box, N the reaction that the table/ground is exerting on the box and Fy the y component of F.

We get that N = |Fp|-|Fy|

By definition we know that the friction force, let's call it Fat, is N*μ

|Fat| = [mg-F*sin(25)] * μ

In this case, checking the forces acting on y, is just needed to get the friction force right.

The block is moving on the x axis, so it has forces acting on it, and an acceleration in that case, the forces acting are Fx (x component of F), and the force of friction.

Fcos(25)-([mg-F*sin(25)]μ) = m*a

The acceleration a is equal to (Fcos(25)-([mg-Fsin(25)]μ))/ (m) [m/s^2]

At this point plug in all the numbers on the calculator and you get what you're searching for, with an human precision.

3. Oct 7, 2014

### paytona

Wouldn't the angle being used be -25* rather than positive??

4. Oct 7, 2014

### Bedeirnur

Why should it be -25?

Isn't it like this

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/10726541_838742206166089_1666516233_n.jpg?oh=83223884bd9c63131013f875e990f880&oe=543720A0&__gda__=1412890032_b43d01452f1154c45db8a98041885387

5. Oct 7, 2014

### paytona

Using your formula I got .568 m/s^2....
Does that sound right?

6. Oct 7, 2014

### Bedeirnur

Well if you've got the solution for that we would know :p

Well, it can be anything, we don't know prior...We just know it isn't negative, as the friction force can block it, not drag it.

7. Oct 8, 2014

### ehild

@Bedeirmuir: You must not give (almost) full solution.

In the problem text, the box is pushed at the angle of 25°. So it is like in the following picture.

ehild

8. Oct 8, 2014

### ehild

Is that angle enclosed with the horizontal or with the vertical?
I see, that 25°was counted from the horizontal. You got Fn rigth.

The acceleration is a vector and its x component satisfies the equation $ma_x=\sum{Fi_x}$ So you need Px and Ffk. No sense to add x components with y components.

ehild