Inclined Definition and 1000 Threads

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 struggling to start this exercice, where I have an inclined air hockey table with an angle alpha. They gave me this chronophotography (this is online) saying that the ratio is 1:1 I really don't know how to proceed since this is online and I'm thinking if I measure this with a ruler...

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.)

Question diagram, attempt at solution below I need to cancel some of the terms in the moment equation but a not sure which ones to start with. I don’t know μ so can not calculate FA, so should probably substitute FA = RB2.

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...

I want to know what are generally the characteristics that can predict or things that could cause the orbit of an object to deviate from zero degrees relative to the equatorial plane of the object that it is orbiting?

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 =...

Now I did what the suggestion said I started my calculation with x(t); x(t) = v0 * cosα * t As I was missing the t component I found it like this t = v0 * sinα/g t = 8,82 s Now I've put t back into x(t) and got this result; x(t) = 441 m Now what I think I have gotten is the complete...

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.

I solved for T on m1 and arrived at 6.72. I plugged that value into the ΣFx equation as shown above (pardon my handwriting) and got a mass of 0.88 kg. The online program indicated that I needed to check my expression for tension, noting that the two tensions are heading in opposite directions...

I understand how the diagram below determined the ##x## and ##y## axis for the velocity vectors but I don't understand the gravity vectors. What I don't understand about the gravity vectors is why is ##-mg## in the ##y-##axis equal to ##-mg\cos\theta## and the ##x-##axis is equal to...

Ateempt of solution: There are two key coordinates in this scenario, the leftmost tip of the rod, which in ##S'## is ##C_{0} = (t', 0, ut',0)## and the opposite tip ##C_{1} = (t', L,ut',0)## An angle ##\phi## could be found through a relationship such as ##tan(\phi) = \frac{ \Delta x}{ \Delta...

I am not sure about my free body diagram. I assume the cylinder rolls without slipping so the forces acting on the cylinder are: Weight directed vertically downwards Normal force directed perpendicular to the plane friction directed upwards, parallel to the plane Am I correct till this point...

Sorry for my English, I'm Spanish. I am not able to make the force diagram for the liquid. I am still in high school and started now little hydrostatic. If someone can give me a light, I would be very .grateful.

This problem was from the chapter on Work and Energy so, I thought of using the principle of conservation of mechanical energy. Clearly, the potential energy of the block decreases by mgh (assuming the block has mass m). This energy should have been converted to kinetic energy, but it clearly...

A single pair of points will be in contact between P and Q. The frictional force will try to make the velocity of these points equal. Say the final angular velocity of Q is ωq. The velocity of points in contact can never be equal because of difference in directions of ωq and ωp. If I break...

i tried to solve the problem with the above way then i calculated Fg= mass x 9.8 = 242.06 N FgII= Fg x sin50 =185.4287N so the force required is 185. is this correct?

Hi, I have a problem with inclined planes. The idea is to calculate the stress in an inclined plane of a bar under tension for which you need the surface. I have no idea how this surface is derived, though. In the attached file, you can see what I mean. For a rectangular cross-section, it's...

I was always a little confused about the rolling down of a body, let's say, a sphere. It's know that to body rotate, from the rest, in a referential frame on the ground [inertial], is necessary a friction, that will just act like a medium that transforms kinetic energy of translation into...

The slope of the v^2 vs h graph is g

I tried solving this by assuming the acceleration of the truck and block to be the same so the block would stay on the incline. Also, I would assume truck ma = static friction, block ma = mgsintheta... then I solved for a to plug into 1st equation to get 12990 N. Is this correct? I wasn't sure...

I think that the answer probably will be yes, but i have a question: Well, with respect to center of mass that's right, but if i choose the axis passing through the contact point, the weight force will produce torque, even without the friction.

Summary:: I have done parts a-d but I can't seem to get e correct! A 6.10 kg block is pushed 9.00 m up a smooth 38.0 ∘ inclined plane by a horizontal force of 78.0 N . If the initial speed of the block is 3.20 m/s up the plane. a. Calculate the initial kinetic energy of the block. (found...

the red line is the initial velocity, the grey parabola is the path of the projectile. hi there...I'm kinda stuck at the part b of this problem. I can do part a with no problem. can anybody explain to me how to do the differentiation needed to solve part b?? by explain I mean explain the...

a. I tried to "rotate" the inclined plane so the surface of the inclined plane becomes horizontal h = Vi sin θi . t - 1/2 g cos ∅ t2 and when it falls to the plane, y = 0 so: 0 = Vi sin θi . t - 1/2 g cos ∅ t2 t = (2 Vi sin θi) / (g cos ∅) Is this correct?b. Particle hits the plane vertically...

Hi Smart People, I have spent a lot of time on this and have looked in Reed's and Wallker's applied mechanics but can't find a clear explanation. Calculating the force required when pulling up an inclined plane whether horizontally to the plane or at an angle seems pretty straight forward...

Hello, I hope I'm in the right topic. I search to know, how many pack the red interlayer on the drawing can support on an easel inclined of 4°. One pack (blue in the drawing) = 2500kg There can be max 20 pack of 2500 kg by side of the easel = 50000 kg One pack measures : 3200x2250x140 mm...

Because the friction is the same in both parts, the calculated acceleration from (b) should be the same for (c) I knew I could find Vf, and thought I could do it with an energy equation Ei=Ef mgh=1/2mv^2 gh=1/2v^2 (2)9.81(1.5)=1/2v^2(2) (square root)29.43=(square root)v^2 v= 5.424 Then...

I determined 42m to be the hypotenuse so I used sine law to find the height of the incline, 10.87m. I used this height in the equation Ei=Ef, since they should be equal. Ei=Ef mgh=1/2mv^2 (at the start there is no kinetic energy, at rest. at the end there is only kinetic, no potential)...

I used work energy theorem between initial top point and point x along the incline(downwards) i got the expression of v then diffrentiated it to get a maxima but it gives me a wrong ans which is 10/6 but the actual ans is 10/3 please tell me what i did wrong

Hi, this is my try: I would appreciate any confirmation or correction . -- Thanks

a=m2g-m1gsinø-m1gcosøu/m1+m2

For the a part of the first problem I did this: mgsinθ < µk*mgcosθ Sinθ < µk*cosθ Tanθ < µk so θ should be less...

Hi If a rigid disc rolls down an incline plane without slipping then the component of weight down the plane causes the disc to accelerate downwards but the frictional force causes a torque which causes the disc to rotate, At the point of rolling without slipping the velocity of the centre of...

This question showed up on my grade 12 physics test. The problem I have with this question is that I did not know the direction that the system would accelerate in, so I just solved as though the mass on the inclined plane would accelerate the system. I expected that if it would accelerate the...

The thing is that my professor said that if the velocity of ##A## is ##400 cm/s##, the velocity of ##B## is ##200 cm/s## because "##B## is connected to two parts of the rope and ##A## is conected just to one part", and he also said that that ##200 cm/s## is the velocity of ##B## with respect to...

I considered the downwards direction and left direction as negative. For ##m_1##, Newton's equations are: ##x) Fr + W_x - T=0## ##y) N - W_y =0## For ##m_2##: ##y) T - W =0## Then, if I replace the data, I get ##T=22.2 N## and then ##m_2=2.2 kg##. With that, for the second question ##m_2=4.4...

Here, I have two doubts 1) if the surfaces are frictionless, then there's no force being applied on the ##x' axis## of ##m_1## except from the weight, so it should be sliding, shouldn't it? So, there's no force that I could apply to ##m_2## to keep ##m_1## in equilibrium since any force that I...

So far I've got E=1/2m(dx/dt)^2+mgh this can be rewritten as (dx/dt)^2=2(E-mgxsin(Θ))/m Would there not be a positive and negative solution? dx/dt=±sqrt(2(E-mgxsin(Θ))/m) or do we discard the negative one? Why would we do so, if that's the case. So far I've only solved the postive solution and...

Homework Statement: A box of mass M=30 kg sits on a ramp which is tilted at angle 13 degrees. The coefficients of static and kinetic friction between the box and the ramp are both 0.12. The box is connected to a spring of spring constant k=9 Newtons per meter. Initially, the box is held in...

FIRST PICTURE I have some doubts here because of the spring... I'll tell you what forces I've drawn. For ##A##, I drew the weight and the force applied by ##B## (the normal force) on the vertical axis; and the elastic force pointing to the right on the horizontal axis. For ##B##, I drew the...

a) Describe the motion of the spool: it will roll down the plane, rotating counter clockwise. I am confused when setting up equation of torque. If I take center of the spool as pivot, only the tension of thread produces torque but the direction will be clockwise and it makes the spool will move...

So here is what is going on in my mind: We have a turn that is 400m away from the center of the turn. The faster the car goes, the harder it is for it to maintain its radius. We have a component of the normal force that points towards the center, and static friction which does the same. I...

So we know that all the energy originates from the spring: E(spring) = (1/2)kd^2 As the block moves up the ramp, friction does work on the block over a distance of 2d: W = μmgcos(θ)* 2d So subtracting the work done by friction from the spring energy, gives us the energy left, so we'll set it...