Incline Definition and 991 Threads

Hi! As written in the summary, I would like to get feedback about my approach to an issue I wanted to explore. In the example I am using cycling, but principally the same theory would apply to similar sports as well. So, the problem was that I wanted to investigate whether the athletes' W/kg...

Hi, a newcomer here. I need to know the value(s) of the force(s) acting on each inclined face of a double ramp. Here's a visualisation of a double inclined ramp (just an example, could be asymmetric): Imagine a body rests on the top faces, let's assume it's a circular bar (a cylinder) which...

FOr this, Use alternate coordinate system With ##ȳ##-axis parallel to incline and ##x̄##-axis parallel to the x-axis. Kinetic energy using this alternate coordinate system is ##T = \frac{1}{2}M\dot x_p^2 + \frac{1}{2}mR^2\dot \phi^2 + \frac{1}{2}m(\dot x̄^2 + \dot ȳ^2 + 2\dot x̄ \dot ȳ...

For this problem, I am confused where they get these two terms from Can someone please explain? I get all the other terms expect those. Thanks alot!

I did some algebra and got the final equation mgh=1/2MR^2omega^2+1/2(2/5MR^2)omega^2, then plugged in the numbers and got an omega value of 111.09. Using 1.7 as h instead of cos(17)*1.7=1.63 I got 114.09. Both answers were wrong. Am I missing something, or did I just screw up the math?

I'm taking college physics without calculus this semester and it's been quite the challenge to say the least. We recently covered free body diagrams and while I understand the different vectors in the FBD, making calculations is killing me. Specifically Newton's 2nd law. The problems range...

For part a and b, I can't see a clear path to finding the answers. In order to find the x component of the applied force I need to know the friction. In order to find the friction I need to find the y component of the applied force, but I can't think of a way to find either. I thought of...

Both myself and my TA gave up, but we found acceleration of the system 9g - 4gsin(30) = 13a a=5.27m/s^2

This is how I tried to do it. The force required to move B up the incline is $kx$ where x is elongation and k is spring constant. we know that spring force is greater than $mg(sin\theta+\mu cos\theta)$. And we can use work-energy theorem to figure out velocity. $0.5*k*x^2=0.5*mv^2$ where...

This was the question: I derived the equations as mentioned in the relevant equations. But I could not solve the equations to find the answer. I realise with respect to inclined plane the acceleration must be a since string cannot slack. With respect to ground, the acceleration of incline is...

a=2/3*g*sin(25*(pi/180))=>a=2.8507 m/s^2 vf=vi+at=>vf=0+2.8507*1.50=>vf=4.2760 m/s So the translational motion of the cylinder is 4.2760 m/s. 4.2760=R*w w=134.04 rad/s PE=mgh=>PE=215*9.8*.108=>PE=227.56 J PE = KE at the end of the roll because of energy conservation. 227.56 =...

For this problem, Why is the tension on each side not equal? For this problem I think the only assumption is that the string is inextensible so the accelerations of the masses are equal. Many thanks!

My attempt: As I need to find acceleration I believe that I need to use F=ma(and thus draw a free body diagram). I drew the block's weight components(mgsinθ, mgcosθ) and concluded that the only force acting on the plane in the horizontal direction is the horizontal component of...

I worked myself into a trigonometry rut. I've tried two approaches, first by not changing the frame of reference, and second by taking the incline as the horizontal x axis. Here is my second attempt: Take the incline as the horizontal. Then the coordinates of target T are: $$ \begin{align} x_T...

Correct answer is (b) 0,620 ms2 My attempt: FBD Equation 1. ∑Fx=0=−ma−Wsin⁡(15)+f N=Wcos⁡15 f=Nμs=379,03N Equation 1. ma=f−Wsin⁡15 15a=379,03−203,1 a=2,20ms2 Equation 2. ∑MG=0=−Nx+f1,5 x=0,75m

Hello everyone! I'm watching this Walter Lewin lecture and am at 5:58 part of the video I'm wondering how there's a frictional torque applied to the cylinder, my reasoning is that the object has forward velocity, and on a perfect cylinder, the slope of the incline touches the cylinder at a...

Ei = 1/2 K (x)^ 2 K = .0152N/m x = .0375 m Ei = 1.06x10^-5 Ef= 1/2mv2 + mgh m = .164kg, v is unknown, h is .0375sin(8.3)=.00541, Ef set equal to Ei 1.06x10^-5=1/2(.164kg)(v^2)+ (.164kg)(9.8)(.00541) v = .3254m/s I have gotten this answer multiple times but it is not correct. I am going...

hey i would like to get some help with this question i try to answer it but it was not correct i like to know what i did wrong and how i can fix it:

##W_{ext}=mgh+KE_f+0=-20(5\sin 37)+(1/2)2(10^2)## ##W_{ext}=-60.18+100=39.81J## But it’s not consistent with ##W=F.d=20*5=100J## I can’t figure it out.

a) $$v=u+at$$ $$v=0+(F/m)t$$ $$v=-10$$ Now $$v^2=0+2as$$ $$s=-5$$ $$W=F.d=(-20)(-5)=100J>40$$ b) $$W=-mgh$$ $$W=-20(3)$$ $$W=-60$$

##W=mgh=100(\sin 37)2=-120J## Right answer! But the question is asking work done by the person. So again I wrote two eqns ##F_N\sin 53+F_D\sin 37-100=10.2a_y## ##F_N\cos 53-F_D\cos 37=-10.2a_x## I just need ##a_x## and ##a_y## to solve.

So for the work done by the kinetic friction, the displacement along the incline is ##s## as given. What I canNOT understand is why the displacement in the y-direction is used for the work done by gravity i.e. ##W = -mgh## where ##h## is the displacement in het y-direction. This instead of the...

But it seems to get too complex and am not sure whether it is correct.. my solution got a bit messy + incorrect Thanks in advance!

Hello, I've worked through the free-body diagram to compute the answer: tan(𝜃) = 0.67 𝜃 = arctan(0.67) = 33.822... The answer is supposed to be approximately 42. Yet, tan(42) is not 0.67, is the suggested answer wrong?

Basically the problem is giving me an initial velocity to start with it goes up the incline before it comes back down. I know how to do everything else in the problem but solve for the initial acceleration up the incline. What would I need to calculate to solve this?

For part 1, I got ## tan \alpha = 1/30 ## ##\alpha = 1.9^{\circ}## ##mgcos(1.9) = 10774N## I'm a little thrown off by the second part. Are we supposed to assume that in the absence of friction, F = N and then substitute F = ma to solve for this?

To find the tension in the rope connecting 6.0 kg block and 4.0 kg block we do 6.0 kg = m1, 4.0 kg = m2, 9.0 kg = M (m_2 + m_1)a - Ma = Mg - m_2 gsin\theta - m_1 gsin\theta Why do we use sin in these equations and not cos?

Summary:: Looking for the formula to calculate force required to push a wheeled cart weighing 227 kg up a 15 degree incline. I’m trying to find the formula for force required to push a 227kg cart with four wheels up an incline that is 15 degrees. From my physics classes I thought the formula...

Hello. Be v grateful any assistance with this. Question is energy to haul 2000kg 4 metres up a 1 in 100 incline with 300 Newtons friction Answer is given as exactly 1.98 kJ. The only figures I can see that relate to anything here are 2000kg minus 10 = 1980, and 1:100 is 0.01 which times...

So I’m having trouble with relative motion with moving inclines and I literally can’t find any help online and my prof does a lot of these problems. This is one of my homework problems, can anyone help me with it please.

Hi, When regarding Gravitational Potential Energy, I know the formula is U=mgh. However, when the object is on an incline (say at an angle of 52 degrees) would it still be mgh or something else? (This isn't homework I simply was just curious).

Hi, I’m wondering if someone can help me understand this question. I can find a resultant force/vector when given an initial angle but I’m stuck here when the only information is the two magnitudes. I think I’m solving for the unknowns but a little lost on how or what equation I should be using...

There are two nonconservative forces in this situation, the work done by the person and the work done by friction - they are the only sources of work that change the total mechanical energy of the mass-Earth system. The initial energy (assuming gravitational potential energy is initially 0) is...

If M moves ##x## along the plane, her height variation in ##x \cos(\alpha)##, and, but I don't know how to find the variation of the height of ##m##

Push-ups take a certain amount of work. If starting from a 45 degree angle instead of from a zero degree angle (the ground), my calculation shows it requires 1/√2 the amount of work. If the relative movement of the body, remaining stiff, is the same, the center of mass moves a certain height...

This is a rough sketch of the model. It is frictionless. I originally tried simple right-angle trig (sin θ =opp/hyp), but that just ends up as sin θ = sin θ , as well as cos θ = cos θ. I feel like there's also a way to manipulate dynamic equations around to equate something that is capable...

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...

First we let the static friction coefficient of a solid cylinder (rigid) be ##\mu_s## (large) and the cylinder roll down the incline (rigid) without slipping as shown below, where f is the friction force: In this case, ##mg\sin(\theta)## is less than ##F_{max}##, where ##F_{CM,max}## is the...

Hi everyone I can't understand how normal force produce torque isn't normal force acts normally on the plane so the perpendicular distance in case if it was like block will be zero, so normal force doesn't produce torque. "case of box"

Why doesn't the incline angle play a role in changing the ##m## component of this equation? ##T = 2π\sqrt{\frac{m}{k}}## FOR QUESTION 25, PART B: ANSWER:

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

My teacher told me that answers are F=1200 N for lifting straight up and F= 360 N for using the ramp. I can get the force for lifting straight up by using 1800=F(1.5)cos(0) but I do not understand why for using the ramp the equation 1800 = F(1.5)cos(72.5) does not produce the correct answer.

Hello! I am stuck on part of a problem and was wondering what I am doing wrong. For part a of the problem, we were asked to find the impact speed. I did this in a photo below given the following values: Θ = 30 degrees. The initial velocity = 10 m/s. The coefficient of kinetic friction = 0.4...

(a) Ridiculously simple though it looks, I can't see how the string will be tight. One of the two has to be true. (1) The static friction ##f_S = mg \sin\theta = 25\times 10\times \sin 30^{\circ} = 122.5\; \text{N} ##. The maximum static friction ##f_S = \mu mg \cos \theta = 0.6\times 25\times...

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...

Figure A It slopes up from Left to Right (0.205911m Rise /4,8m run=4.28%)]

I think it will be between 11 and 12 degrees but don't know how to re-arrange the equation or if I was even using the correct method

Good afternoon everyone, I need help with a certain calculation. How much power is needed to pull a 250lb load up a 45 degree incline at 5 mph. I am working on a project at school and would like to know what kind of motor we can use. Thank you very much!

Hello I have a problem with calculating the support reactions for a beam. Lefts side of beam has a pin connection so it takes both Fx,Fy. Right side of the beam has a horizontal roller and it takes only Fy in the direction of the wall. Therefore at the pin support Fy=9kN, but how do i figure out...