What Is the Required Angle for a Block to Slide Up an Inclined Plane?

[mxtiger67]

You have a block on the incline plane 130 Newtons and it's attached by a rope through a "frictionless" pulley to another block of 45 Newtons hanging of the side of a plane. The question is what angle pheida has to be in order for a 130 Newtons block to start sliding up the plane (not accelerating) when coefficient of friction is 0.62.
I think the equation starts like this:
45=130*sinO + 0.62*130*cosO
I've tried solving for it, but it's not coming out.
Can anyone help!

 

[derekmohammed]

One thing to remember is that When you are drawing your free body diagram draw it so the X-axis is parrel to the inclined plane, this makes it easier to solve the equations.

Since the box will be accelerating at a constant Velocity it will be the same as if the two boxes were in equilibrium and the Fnet in the X will be Zero.

ThereFore: Fnet= Fgsin(Angle) + Ffriction - Tension Force
FNet=0
Therefore: Ff+ 130sin(angle) = Tension Force sin(angle)
(If you are wondering why the trig function is sin and not cos it is because we are using the inclined plane as the X-axis.

I think that these hints will help

 

[mxtiger67]

How do I find Tension? Is it just the weight of the second block, 45Newtons?

 

[Pyrrhus]

You apply Newton's 1sd Law

$$\sum_{i=1}^{n} \vec{F}_{i} = 0 \rightarrow v=const$$

Assuming massless-frictionless pulley:

The forces acting on the object on hanging it's the tension and the weight, and in the object on the incline, the normal force, the weight, the tension and friction

For the object on the incline:

$$n = m_{incline}g \cos \theta$$

$$T - m_{incline}g \sin \theta - \mu m_{incline}g \cos \theta = 0 (1)$$

For the object hanging:

$$T - m_{hanging}g = 0$$

$$m_{hanging}g - T = 0 (2)$$

Ok i will do the math (it's simple trigonometry):

$$m_{hanging}g = m_{incline}g \sin \theta + \mu m_{incline}g \cos \theta$$

Rewriting because we got the weights:

$$W_{hanging} = W_{incline} \sin \theta + \mu W_{incline} \cos \theta$$

Dividing by $W_{incline}$ and $\cos \theta$

$$\frac{W_{hanging}}{W_{incline}} \frac{1}{\cos \theta} = \frac{\sin \theta}{\cos \theta} + \mu$$

Using Trigonometry identities $\frac{\sin \theta}{\cos \theta} = \tan \theta$ and $\frac{1}{\cos \theta} = \sec \theta$

$$\frac{W_{hanging}}{W_{incline}} \sec \theta = \tan \theta + \mu$$

Squaring both sides

$$\frac{W_{hanging}^{2}}{W_{incline}^{2}} \sec^{2} \theta = (\tan \theta + \mu)^{2}$$

$$\frac{W_{hanging}^{2}}{W_{incline}^{2}} \sec^{2} \theta = \tan^{2} \theta + 2 \mu \tan \theta + \mu^{2}$$

Using trigonometry identity $1 + \tan^2 \theta= \sec^2 \theta$

$$\frac{W_{hanging}^{2}}{W_{incline}^{2}} (1 + \tan^2 \theta) = \tan^{2} \theta + 2 \mu \tan \theta + \mu^{2}$$

$$\frac{W_{hanging}^{2}}{W_{incline}^{2}} + \frac{W_{hanging}^{2}}{W_{incline}^{2}} \tan^2 \theta= \tan^{2} \theta + 2 \mu \tan \theta + \mu^{2}$$

Rearraging:

$$(\frac{W_{hanging}^{2}}{W_{incline}^{2}} - 1) \tan^2 \theta -2 \mu \tan \theta + \frac{W_{hanging}^{2}}{W_{incline}^{2}} - \mu^{2} = 0$$

Now remember $Ax^2 + Bx + C = 0$

The solutions are

$$\theta_{1} = -50.8491231^{o}$$

$$\theta_{2} = -17.58249194^{o}$$

 

[mxtiger67]

Yeah and both of those posts are just rewritten formulas of what I've already wrote. I'm having problem with the math trying to figure the formula out. I keep coming up with angle of 48.7 degrees, but when I try to sub it and check, it's coming out wrong.

 

[Pyrrhus]

Yeah and both of those posts are just rewritten formulas of what I've already wrote. I'm having problem with the math trying to figure the formula out. I keep coming up with angle of 48.7 degrees, but when I try to sub it and check, it's coming out wrong.

I edited my above reply with the math

 

[mxtiger67]

So basically from the angles that you came up with, such angle doesn't exist. I did the math a bit differently and my angles came out differently, but still are not working.

I have
45=130sin(angle)+80.6cos(angle) divide by 45
1=2.89sin(angle)+1.79cos(angle) now put 2.89sin(angle) on the other side and square
1-5.78sin(angle)+8.35sin*2(angle)=3.2cos*2(angle) sub (1-sin*2(angle) for cos*2
1-5.78sin(angle)+8.35sin*2(anlge)-3.2+3.2sin*2(angle)=0

angle is 48.84 and -14.7
now when I sub this angle back in the equation to check the answer, it doesn't work.
IT'S DRIVING ME CRAZY, AND I CAN'T FIND ANY MISTAKES!
Maybe you can help?

 

[Pyrrhus]

You saw the math, those are the solutions, plug them in your equation and they will work, you can change them to positive by adding 180 and such methods.

By the way i finished your method, it gives the same solutions as mine, as expected.

Here is the equation:

$$1 - \frac{W_{incline}^2}{W_{hanging}^2} \mu^2 - 2 \frac{W_{incline}}{W_{hanging}} \sin \theta + (\frac{W_{incline}^2}{W_{hanging}^2} + \frac{W_{incline}^2}{W_{hanging}^2} \mu^2) \sin^2 \theta = 0$$

Solutions:

This one doesn't works:
$$\theta_{1} = 17.58249205^{o}$$

This one works:
$$\theta_{2} = -50.84891189^{o}$$

 

[mxtiger67]

ok, then my mistake is somewhere where I solve for sin(angle)
Ax^2 + Bx + C = 0, I get

-(-5.78)+or-sqrroot(5.78*2-4(11.55)(-2.2))/(2(11.55))
so I end up with sin(angle)=.752 or -.25
did I put numbers wrong in the equation?
because I still get 50 and -15? where am I wrong?
and after I sub'ed your answers into an original equation, it doesn't work, unless again, I'm doing something wrong.

 

[Pyrrhus]

Work out the algebra then substitute the values, if you substitute the values first it will be messy, like you have seen, so do it with letters then substitue the values, like i have done above. Also it will be easier for me to check your work.

 

[mxtiger67]

$$sin \theta = -(-2 \frac{W_{incline}}{W_{hanging}} ) +or - square root of (2 \frac{W_{incline}}{W_{hanging}} )^2 - 4(\frac{W_{incline}^2}{W_{hanging}^2} + \frac{W_{incline}^2}{W_{hanging}^2} \mu^2)(1 - \frac{W_{incline}^2}{W_{hanging}^2} \mu^2)/(2(\frac{W_{incline}^2}{W_{hanging}^2} + \frac{W_{incline}^2}{W_{hanging}^2} \mu^2)$$
I think this is more confusing, than just using the numbers.

 

[Pyrrhus]

Sinc you are new to latex, i will just type it in my work (using the method you debscribed), if you want finish yours, so you learn how to use Latex

$$W_{hanging} = W_{incline} \sin \theta + W_{incline} \mu \cos \theta$$

$$1 = \frac{W_{incline}}{W_{hanging}} \sin \theta + \frac{W_{incline}}{W_{hanging}} \mu \cos \theta$$

$$1 - \frac{W_{incline}}{W_{hanging}} \sin \theta = \frac{W_{incline}}{W_{hanging}} \mu \cos \theta$$

$$1 - 2 \frac{W_{incline}}{W_{hanging}} \sin \theta + \frac{W_{incline}^2}{W_{hanging}^2} \sin^2 \theta = \frac{W_{incline}^2}{W_{hanging}^2} \mu^2 \cos^2 \theta$$

$$1 - 2 \frac{W_{incline}}{W_{hanging}} \sin \theta + \frac{W_{incline}^2}{W_{hanging}^2} \sin^2 \theta = \frac{W_{incline}^2}{W_{hanging}^2} \mu^2 (1 - \sin^2 \theta)$$

$$1 - 2 \frac{W_{incline}}{W_{hanging}} \sin \theta + \frac{W_{incline}^2}{W_{hanging}^2} \sin^2 \theta = \frac{W_{incline}^2}{W_{hanging}^2} \mu^2 - \frac{W_{incline}^2}{W_{hanging}^2} \mu^2 \sin^2 \theta$$

$$1 - \frac{W_{incline}^2}{W_{hanging}^2} \mu^2 - 2 \frac{W_{incline}}{W_{hanging}} \sin \theta + \frac{W_{incline}^2}{W_{hanging}^2} \sin^2 \theta + \frac{W_{incline}^2}{W_{hanging}^2} \mu^2 \sin^2 \theta= 0$$