Calculating Forces on a Crate Moving up a Ramp

[MysticDude]

Homework Statement

In Fig. 5-39, a crate of mass m = 100 kg is pushed at a constant speed up a frictionless ramp (\theta = 30.0°) by a horizontal force \vec{F}. What are the magnitudes of (a) \vec{F} and (b) the force on the crate from the ramp?

Here is the image:
[PLAIN]http://img42.imageshack.us/img42/1110/physicsnum24.jpg

Homework Equations

Magnitude of normal force: F_N = m(g + a_y)
a = 0(constant speed)

The Attempt at a Solution

a. 565N
b. Attempt at a: Well since F = 0 since F = ma and a = 0, Fcosθ and mgsinθ have to cancel each other right?
So I'm thinking that it should be Fcosθ - mgsinθ = 0. In other words, F = \frac{mgsin(\theta)}{cos(\theta)}. Which would give me 565kg.<br /> <br /> <b> Attempt at b: </b> I&#039;m thinking about using mgsinθ where θ = 120°. That would only give me 849N. <br /> <br /> Thanks for any help! Please point out any flaws in my logic with these problems :D

 

[ThyQuotidian]

 

[MysticDude]

Can you please evaluate your drawing please? Like what's the n in the second equation? You just taught me a lesson too. I really need to start drawing free body diagrams lulz. :D

 

[ThyQuotidian]

Oh yeah sorry haha. Well n is the normal force because it's always perpendicular to the surface. Fa is force applied and Fg is the force of gravity.

Looking at the summation of Fx you can easily solve for Fa by adding both sides by Fgsintheta and dividing costheta.

Fa = Fgsinθ/cosθ

The reason the summation of Fx = 0 is because there is no acceleration, the problem stated that there was a constant speed.

The force that the ramp exerts on the crate is the normal force (the force that makes the crate not go through the ramp)

So you want to solve for n with the summation of Fy equation.

n = Fasinθ + Fgcosθ

 

[MysticDude]

okay so after I get F_a I just plug it into the second equation to get n, right?

 

[ThyQuotidian]

Look at my last post, I edited it. But yeah that's right.

 

[MysticDude]

I don't think I'm doing this right. I am supposed to use θ = 30 right? If so, then I should use F_g = 9.8*100 and sinθ = sin(30°) and cosθ = cos(30°). If I use that, I get F_a = 565N. Okay after that I will use your second equation and plus in 566sin(30°) + 980cos(30°) to get 1132N. Is this what you get?

Note: I'm rounding off for simplicity. I don't want to use chains of numbers on the forum now :D.

 

[ThyQuotidian]

I don't think I'm doing this right. I am supposed to use θ = 30 right? If so, then I should use F_g = 9.8*100 and sinθ = sin(30°) and cosθ = cos(30°). If I use that, I get F_a = 565N. Okay after that I will use your second equation and plus in 566sin(30°) + 980cos(30°) to get 1132N. Is this what you get?

Note: I'm rounding off for simplicity. I don't want to use chains of numbers on the forum now :D.

Yep it looks like you did it right. Think about it. 100 kg isn't light. Not just that but you're pushing against the ramp you're not pushing at an angle so that makes the force that the ramp feels go up.

***100kg is 220 lbs!***

 

[MysticDude]

Okay I get it, but where do you get the logic to make to use these equations? That is my only problem in physics. The logic. In other words, did you know how or do this problem because of experience or you just knew your formulas?

 

[ThyQuotidian]

Where I always start is by drawing a picture of what's going on. Then I start drawing a free body diagram. The reason I tilted the axis is because you always want your x-axis or y-axis parallel to the motion.

For instance the box was going up the ramp so I decided to let the x-axis be parallel to which way the box was going.

Next I add the forces. I always start with gravity then normal..then applied and then all the other forces like friction and what-not.

To get the equations you have to ask yourself...what's affecting the x-axis? Well looking at the free-body diagram the normal force isn't affecting the x-axis...however gravity is and so is the applied force. But not 100% of the normal force OR gravity so you have to take the x component of each of those. Since the x component of Fa is in the positive area and the x component of gravity is negative you subtract Fg from Fa (Fa - Fg) and that has to equal some (ma) because there's an acceleration going that way.

In your book it should have (Fnet = ma)

So the way I think of it is...The sum of all the forces = 0 OR some mass times acceleration

I did the same thinking for the y-axis.