Inclined plane Definition and 812 Threads

At 3m: U = 3 x 9.8 x 80 = 2352J U at 6m = 6 x 9.8 x 80 = 4704J k = 1/2mv^2 (4704-2352)J = 1/2 (80) v^2 v = 7.67 m/s At 0m (bottom): U = 0 4704J = 1/2 (80) v^2 = 10.84 m/s Okay so what is bothering me here is just that my working doesn't take into account the...

This isn't really a question per se, but it recently just 'clicked' for me and I would like to share what made that do so. There are dozens of threads answering this question, but in my opinion most (but not all; there are some good threads on this forum, I just want to show how I came to...

I think that both kids experience the same acceleration (irrespective of mass) since the only force pushing them downwards is acceleration due to gravity, which is the same for both of them. Thus, since they start sliding down the hill at the same time (assumption), and are accelerating at the...

I'm not really asking for a solution for this problem I just want to clear up a confusion I have. Why are they multiplying the weight by the sin and cosine of the 30-degree angle? Isn't weight not affected by anything since it's constant? Also is the angle of friction 0 because it's a...

I am a bit confused with velocities in this problem. From Philipp's view, the coin's initial velocity is zero, so its transfer kinetic energy is also zero. When I am standing on a non-moving ground, is the coin's initial velocity ##v## in direction the walkway is moving? But won't I get then...

I want to use the Lagrangian approach to find the equation of motion for a mass sliding down a frictionless inclined plane. I call the length of the incline a and the angle that the incline makes with the horizontal b. Then the mass has kinetic energy 1/2m(da/dt)2 and the potential energy should...

The virtual displacement should be given by $$ \delta\vec{r} = \begin{pmatrix} \cos(\alpha) \\ \sin(\alpha) \\ \end{pmatrix} \delta s $$ where ##\delta s## is a displacement parallel to the plane. The relevant force should be the gravitational force, as given above. Thus, the equations of...

a = 9.8*sin(25*pi/180)=>a=4.1417 m/s^2 vf^2=vi^2+2*a*s=>vf=sqrt(0^2+2*4.1417*3)=>vf=4.9850 m/s Meanwhile the correct answer is: (vf+vi)/2=>(vf+0)/2=2=>vf=4 m/s Why is my answer wrong? It seems that the acceleration is what is wrong, but I don't understand why.

For part (c) of this problem, My working is However, the tricky part is to find theta. I tried to draw the situation so that I could find theta: It appears that theta = 90 degrees. However, this does not seem to be correct. Does anybody please know how to correctly find theta in terms of...

I was looking at an example of fluid mechanics and I don't understand this. Statement figures: CONTINUITY EQUATION $$\left. \dfrac{dm}{dt}\right]_{MC}=(\dot{m}_2+\dot{m}_3)-\dot{m}_1=0$$ $$\dot{m}_1=\dot{m}_2+\dot{m}_3$$ $$\rho c_1A_1=\rho c_2A_2+\rho c_3A_3$$ $$\rho c_1 h1=\rho c_2 a1+\rho...

So i am tried to conserve momentum and use conservation of mechanical energy but won't there be psuedo force acting on the block if i am solving from non inertial frame ?. If i ignore the pseudo force and simply use C.O.M.E and include the K.E of the wedge and solve normally i do get the...

This is a homework problem of my grand daughter. The question is to find out the conditions of an object M on a slope with angle shown and applied force "F". I find there are 3 conditions, sliding up, sliding down and not moving. This is my work. I just want to get comments on my work: At the...

I have seen a few posts on this subject before, but none have really answered my question. For clarity, I will refer to the 1st example as a wedge, and the second as a ramp (although both are of course inclined planes). With both examples that I outline below, we will assume no friction, and a...

I was thinking about how various objects would slide down on an inclined plane, and I just couldn't figure this problem out. So let's say I have this screw or cone on its side, on an inclined plane. If friction exists, what would the motion of the screw be as it slides down the inclined plane...

Suppose a square lattice. The planes are such as the image below: I light wave incides perpendicular to the square lattice. The first maximum occurs for bragg angle (angle with the plane (griding angle) as ##\theta_B = 30°## (blue/green), green/blue in the figure). The angle that the...

The best I could do was draw a forces diagram. I know that friction would be working up when the block is on the point of slipping down the plane and friction will be acting down the slope against the direction of motion when the block is on the point of slipping up the slope. (not even sure if...

hey i would like to understand something, i solve this question but i don't understand why my answer is right, first of all we learn that in problems like this we need to disassemble into components the mg and that what i try to do here but i didnt get the right answer so then i try to do same...

When we take the x-axis parallel to incline surface its clear that the horizontal component of weight is causing the block to come down but when we take the standard orientation its not so clear to me. Is horizontal component of ##F_N## causing the block to come down? <Moderator's note: Use of...

If on a flat ground, we exert a force F to move forward, then we go to an incline plane of theta degrees. Why wouldn't the force F2 to move up the incline plane with respect to ground be F2*cos(theta) = F --> F2 = F/cos(theta) disregarding the effects of gravity?

For lower half ,$$Fnet=-\mu F_N+mg\sin \phi$$ For upper half, $$v^2=u^2+2as$$ (s is half of the total slant distance) $$v^2=0+2\frac{mg\sin \phi}ms$$ $$v=\sqrt{2g\sin \phi s}$$ again for lower half, $$v^2=u^2+2as$$ $$0=2g\sin \phi s+2\frac{-\mu F_N+mg\sin \phi}ms$$ $$\mu=\frac{2gm\sin...

1) By the Work-Energy Theorem, ##W=K_f-K_i=\frac{1}{2}I_{0}\omega^2=\frac{L^2}{2I_0}.## 2) By assuming that the initial length of the spring is ##0##, calling its final length ##S## and ##T## the tension in the rope connecting the pulley and mass ##m_p## I have: ##\begin{cases}(kS-T)r=0\\ m_p...

1. ##-f_k\cos\theta-T\cos\theta+F_n\cos\alpha=m_2a_x## 2. ##f_k\sin\theta+T\sin\theta+F_n\sin\alpha-m_2g=-m_2a_y## 3. ##T-m_1g=m_1a_y## I am unable to get anywhere. There are accelerations in x , y directions. I need the value of acceleration. Then I can simply use ##s=ut+\frac12at^2##

Body A rests on a inclined plane of body B . the angle of slope is α , the coefficient of friction between the two bodies is μ . Body A does not slip on body B because we accelerate body B with a. What is the minimum and maximum acceleration required for body A not to slip? What will be the...

It's the body. So there's friction on that plane and there's tension also. $$L=\frac{1}{2}m_1\dot{x}^2+\frac{1}{2}m_2\dot{x}^2-m_2g(l-x)-m_1gx\sin\theta$$ $$f=\mu N=-\mu m_1 g\dot{x}\cos\theta$$ I had found the frictional force's equation from [the...

The acceleration and velocity of a body rolling down without slipping on a frictionless inclined plane are given by $$ a=\dfrac{mg\sin \theta }{m+\dfrac{I}{r^{2}}}=\dfrac{g\sin \theta }{1+\dfrac{K^{2}}{r^{2}}} \cdots(1) $$ $$...

(I drew motion in the opposite direction so the object would rotate trigonometrically but it should be the same thing) I have just finished the Kinetic Energy and Work chapter in my course and this is the last problem from the problem set. I have not worked many problems with the Work-Kinetic...

I’m pretty sure that the force on the sphere by the wall and plane has to equal mg so the sum of the normal force is steered by the wall and plane has to equal mg. I’m not sure where to go after this. Is mg the answer or is there something I’m missing?Here is Fig: 4-31:

##V_x = V_0 cos \theta ## ##x = V_0 cos \theta t## ##V_y = V_0 cos \theta ## ##y = V_0 cos \theta t## ##F_x = m\ddot{x}## ##-mgsin \phi = m\ddot{x}## ##\dot{x} = -gtsin\phi + V_x## ##x = -\frac{1}{2} gt^2 sin \phi + V_x t## ##x = -\frac{1}{2} gt^2 sin \phi + v_0 cos\theta t## ##F_y =...

I apologize if this is the wrong area to post this in, I've never posted on thisforum before. I'm trying to form a FBD of a mechanism that uses linear motion to actuate a pin in a transverse direction. I've attached the general idea in a picture where the pin is free to move up and down. In...

In my textbook, it is stated that "if an object elastically hit an frictionless inclined surface with angle between the vector of initial velocity and an imaginary line that is perpendicular to the surface ##\alpha##,then the angle between the line and final velocity vector will also be...

ma = mg * sinα - fmg * cosα a = g (sinα-f * cosα) v = g*t(sinα - f * cosα) 14.7 = 10 * 2 (sin60 - f * cos60) f = 0.26 Can someone please check if my solution is correct? I'd really appreciate that.

This image was provided, I've completed the first part of the question and got a = 4.8m/s^2 as well as T1= 24.5N and T2=34.3N. not sure about my answers though. also I don't understand the mass in static equilibrium part, can anyone explain how to solve that? Thanks.

a) When the system is in motion for the first time, the force causing ##M## to move is contact force with ##m## so: $$\Sigma F=M.a$$ $$N \sin \alpha=M.a$$ $$mg \cos \alpha \sin \alpha =M.a$$ $$a=\frac{mg \cos \alpha \sin \alpha}{M}$$ Is that correct? b) Is acceleration of ##m## the same as...

Confused how to approach problem after making a free body diagram and finding the length of the inclined plane

This was the answer key provided: My questions are the following: if the force required for rotational equilibrium is more than the limiting static friction, then the body will rotate aka slip over the surface. When it slips, the frictional force will be kinetic and not static, right? If I...

I have calculated the force equation on the x-plane which is f - mgsin(30) - friction force = ma and from the equation vf = vi + at resulting a = v/10

Summary:: Please see the picture below Let say: ##W_1## is weight of ##m_1## ##W_2## is weight of ##m_2## ##f_1## is friction on ##m_1## ##f_2## is friction on ##m_2## I want to find the acceleration of the system. Since I don't know in which direction they will move, I just assume ##m_1##...

6.2(9.8)=60.8 (incorrect) 46.2cos40.4= 35.2 (incorrect)

Hi, I'm missing something really stupid here... The problem is the usual one with a block sliding down (or moving up, it should be the same) a frictionless inclined plane,which itself is free to move on a orizontal frictionless surface. These problems are usually solved stating that only...

I did part a and part b but stuck in part c. Could you help me? (Part a and b attached.)

Hello there, I have a question regarding this problem. I have no problem with part A. However, in part B, my solution manual states that the hollow cylinder will reach the bottom last. Why is it? I mean shouldn't the solid cylinder and the hollow one reach the bottom at the same time? you know...

Hey Everyone, my physics teacher has assigned us a task which involves predicting the range of a ball falling down an inclined plane into a free-fall, the equation for the final velocity of the ball down the ramp, accounting for rotational velocity has been provided, this is the initial velocity...

Hello! So my main and first problem about this question is, I do not know what the problem is about. What I mean by that is, in class we talked about pendulums and are given formulas and assignments regarding pendulums. But this problem here does not seem like it has anything to do with...

Hello! So the way I have tried to solve this problem is the following;Since it is an inclined plane and the cofficient of static friction is known, getting to the angle at which the box starts sliding is the following ##μH = \frac {sin (\alpha)} {cos(\alpha)} = μH = tan(\alpha) ## ## \alpha =...

So after trying to calculate the horizontal forces to solve it: f + Wx(gravity force component of x) - Fy ( the Force that is supposedly giving the the acceleration) = 0 It got to me that the question said "plane has the acceleration" is that even possible? Unless the plane is another object...

A block of mass 0.2 kg which slides without friction on a θ = 30° incline is connected to the top of the incline by a mass-less spring of relaxed length of 23.75 cm and spring constant 80 N/m as shown in the following figure. (a) How far from the top of the incline does the block stop? (b) If...

I have problems to even start with this exercise.