Forces Definition and 1000 Threads

Question 1: I believe that the ratio would be b. 8:1 because by combining the formula for kinetic energy and momentum the expression Ek=p^2/2m can be obtained. Thus, for a body of mass 2kg with twice the momentum: Ek=2^2/2*2=1 For a body of mass 4kg with half the momentum: Ek=1^2/2*4=1/8...

Letting the normal force on the pad on the left be 1 and the one on the right be 2, Normal force R1 + R2 = W Since the sled is not rotating, net moment =0, Taking moment about the right pad, W(ℓ/2 + x) - Hh - R1ℓ =0 R1 = W/2 + Wx/ℓ - Hh/ℓ Since the rate of the pads being worn is...

I have attempted this problem by solving for the normal force. (16.73)(9.8)-51.25(sin)(49.1). I tried to work that out but it was incorrect because apparently the vertical force is zero. Could I get an explanation on what that means and where I should start?

Hello: I was looking for a widespread convention (akin to Hibbeler's, Beer's, etc) that deals with the sign convention of a vertical bar for bending moments. For example, without knowing in advance, how do I draw the bending moment at a cut passing through point E in the figure attached? Beam...

Here is my solution I just need it to be confirmed Many thanks in advance!

Could I please ask for help regarding the following question: D, E and F are the midpoints of the sides QR, RP and PQ respectively of triangle PQR whose circumcenter is O. Forces of magnitude kQR, kRP and kPQ act at O in directions ##\overrightarrow{OD}##, ##\overrightarrow{OE}## and...

I am also attaching picture of the figure as well as my diagram to showcase the forces. I have devised these equations because as per the question, r(A) is acting like a reference. The constraint r(A)+r(B)=l--------(i) transforms to r''(B)= - r''(A) ------- (ii) Using- F(A)-T=m(A)*r''(A)...

I am sure that the answer will be negative, but I'm not 100% sure on how to solve the problem. My initial answer was -350 J. Please help. Thanks!

in the limit as Re→0 , we can neglect the material derivate of v ( Dv/Dt =0 ) but why in books they always make the gravity effects equal to 0? i can't understand and no one really explains this stuff

Could I please ask for help with the following question: A lamina is in the shape of an equilateral triangle ABC, and D, E, F are the midpoints of BC, CA, AB respectively. Forces of magnitude 4N, 8N, 4N, 3N, 3N act along AB, BC, CA, BE, CF respectively, the direction of each force being...

Given: r1 = 0.75 r2 = 1.5 Ok, i have done statics and dynamics, but this book does things differently than the way i accomplished it in the previous classes. This is for mechanical engineering design class. Here is my question. They take sum of the moments along X, which goes through the center...

Hello! To this I did what was recommended and this is what it looks like $$ F = mg $$ $$ F = \rho * V * g $$ $$ F = \ rho * \pi^2 * h * g $$ Than for the surface tension I did the same thing to get an expression for F. $$ y = \frac {F} {2 \pi r}$$ Than tried to get F out and than...

Hi everyone, I have a question that I am struggling and need your help. So I am working on a project which is essentially two syringes, A & B, with different cross-sectional areas, A1 & A2, connected via a tube filled with water. If we assume loss-less system, Pascal's law, the input force...

So I already know the normal force is 268.08 from a previous part of the question. I thought that the friction force must be less or equal to uFN for an object to stay in static equilibrium. So Tcos(angle)=uFN T=uFN/cos(angle)=116.49 But the answer is suppose to be 133.37.

I refer to the website below (for more information): http://www1.lasalle.edu/~blum/p106wks/pl106_Pendulum.htm#:~:text=The forces acting on the,the tension of the string.&text=The net radial force leads,is v2/r.) P.S. I'll insert my specific questions in the following paragraphs in this format...

ma=-k2l+k1l (ma-k1l)/l=-k (-65.12-35.7)/0.51=-k k=197.69, but answer should be 57.7

So I know the acceleration is 0, so the net force is 0. QE=1.6E-3*9.9E3 k hat = 15.84 k hat (thats one force) qv x B = q(v i hat + v j hat) x (0.51 i hat + 0.10 j hat) =q(0.10 v k hat- 0.51 v k hat) =q(-0.41 v k hat) =-0.000656 v k hat Now solving for velocity, 0.000656v=15.84...

So first I did the vector stuff. r2-r1= 1.3 i hat-47.5 j hat-14.5 k hat magnitude = 49.68 magnitude squared = 2468.19 Now plugging it all in... F=9E9*6.3E-3*2.8E-3/2468.19=64.322 y vector, -47.5/49.68=-0.956119 j hat Multiply this by force and I get -61.499 but answer should be -36.14

So, I decided to do Fg=ma I tried 25(9.8)=1.62m m= mass However, when i did it i got a mass of 151 but when i put it into its separate equations, the fg is not 25x larger so fg=ma fg=151(9.8) = 1479.8and then fg=ma fg=151(1.62)= 244.62 but 1479.8/244.62 is not 25 it ends up being 6.05...

I know its something to do with my unit conversions because my answer is the right numbers but too small by a factor of 10000. I convert the radius to meters by multiplying it by 1000. I convert the density to kg/m^3 by dividing it by 10. I find the volume using the equation (4/3)*pi*radius^3...

So what I did was find each of the forces the masses had on m1 using the above equation. From m2 I found 19.975 in the negative i hat and for m4 i found 29.96 in the positive k hat direction using the above equation. For m3 I used pythagorean theorem to calculate r, which was 3.25 (so r^2 was...

So 0=q1q3/r+q2q3/(13.6-r) 0=-8.5/r-3.63/13.6+3.63/r 0.2669=3.63/r-8.5/r r=-18.2465, but the answer is supposed to be 8.24

Hello, Trivial question: a system is isolated and all its internal forces are conservative. Because of Newton's 3rd law, all internal forces are pairwise and the net internal force is always zero (regardless of the forces being conservative or not) hence the system's total momentum is conserved...

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Don't know how to approach this problem. I know how to solve 2 opposite constant forces, but in this problem both forces (water and mercury) change with time, until reaching equilibrium.

Hello, I would like to review and validate some concepts that I have been recently thinking about. Hope this is correct and useful to others that need to refresh these concepts. Forces can be classified as either conservative or nonconservative. Dissipative forces are always nonconservative...

The system shown is in equilibrium, Find the forces in the AB, BE,ED and BC branches.

I know that the force must be a central force and that under central forces, angular momentum is conserved. But I am unable to mathematically show if the angular and linear momentum are constants. Radial Momentum ##p=m\dot r = ma\dot \theta=ma\omega## Angular Momentum ##L=mr^2\dot\theta =...

We have 2 forces affecting the rope: 1. Gravitational force of the body ##=mg## and 2. Force of air = Force of drag= ##F_{AIR}##. The length of the rope is shortening with the velocity ##v_k##. So to figure out the angle ##\theta## I wrote: ##R##= force of rope ##R_x = F_{AIR}## ##R_y = mg##...

Can anyone help me solve the magnitude and direction of force at C. I've managed to get a value for AB. Using basic trig and substituting into F= ((9x150+19.62×350) / 50) / cos9.782 Therefore producing an upward force at AB=166.7N when it comes to force @ C.. I'm lost!

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...

My first question is actually, what happens when any two objects get near each other? This question is often phrased as "Why can't you really touch anything?" or "Why can't you walk through walls?" I have heard two answers: 1. the repulsion between electrons 2. the Pauli exclusion principle...

Can anyone please help me with the following? Three forces which act along the sides AB, BC and CD of a regular hexagon ABCDEF of side 2a, have a resultant which acts along DF. When a couple of 4Pa in the sense CBA is added in the plane of the hexagon, the resultant acts along CA. Find the...

I think I've made a mistake with the vectors. I was trying to get to the same point as the original working (made by IWTSE, not me), but one of the sides of a triangle can't be 0. I would appreciate some help.

The answer is 280N but I don't have any idea hoe to get to the anseer. Atleast a hint would help.

Help, this one is completely different, the weird pulley is making it hard for me and I don't even know what to do with it and where to start. The answer is 185N but I want to know how.

Hi I need help with part d. I think for part q(i) the force is horizontally and constant and equal to what I have found in part c. and for part d(ii) the force is upwards and equal to what I have found in part b.

Hi I'm very stuck on what to do for these 2 questions I got wrong Can someone please help me on what triangle I need to sketch out in order to find the 2 components of forces for F1 and F2. I'm assuming you have to make use of the sine or cosine rule here I'd be grateful for any help given...

I have attempted and got 2 different solution 1: 2:

The ansatz for the 5D metric is \begin{equation} G_{\mu \nu}= g_{\mu \nu}+ \phi A_{\mu} A_{\nu}, \end{equation} \begin{equation} G_{5\nu} = \phi A_{\nu}, \end{equation} \begin{equation} G_{55} = \phi. \end{equation} This information was extremely enlightening for me, but what's the analogous...

Resolving the weight of the cylinder c, i get Mgcosθ (-y) and Mgsinθ (-x) mgsinθ - Fs - T = ma ---(1) (where Fs is frictional force and T is tension) τ = I α (where τ is torque and α is angular acceleration) torque is produced by both tension and frictional force (T-Fs) * r = 0.5 m r^2 α...

Could I please ask for advice with the following: ABC is a right-angled triangle in which AB = 4a; BC = 3a. Forces of magnitudes P, Q and R act along the directed sides AB, BC and CA respectively. a) Find the ratios P:Q:R if their resultant is a couple. b) If the force along the directed line...

Could I please ask for help with the following: ABC is a right-angled triangle in which AB = 4a; BC = 3a. Forces of magnitudes P, Q and R act along the directed sides AB, BC and CA respectively. Find the ratios P:Q:R if their resultant is a couple. Book answer is 4 : 3 : 5 Here's my diagram...

Here's a diagram of the object My Question: My friend and I were solving this problem. We both found Fxnet and Fynet. However to find work I did W= Fnet*d whereas he did W= Fxnet* d. He claims that since it's sliding on the floor it can only go in the x-direction, and it can't go in the...

Can I please ask for help with the following: Forces proportional to the sides of a quadrilateral taken in order act respectively along those sides. Prove that the resultant of the system is a couple whose magnitude is represented by twice the area of the quadrilateral. Not sure where to start...

As far as I know, the grand prize of a Theory Of Everything is mathematically uniting of all forces in the conditions close to the big bang but one of the main problems from the GR end of things is that gravity is not actually a real force to be combined with anything. All the most popular...

From this question, I do not understand why there are three forces exerted at Point C (2 of it being the tension by weight A and the other is the tension by weight B) I understand that there is tension by the two weights but why is there 2 forces exerted by weight A at point C? From the...