Another physics problem....Not my fav. subject?
920693
For equilibrium of the board-and-boulder system, two conditions must be satisfied:\n\n!) Sum of vertical forces must be zero and \n\n2) The turning effects of the forces of Will and Cliff (about the center-of-mas of the system) must total up to zero.\n\nNow considering the vertical forces: \nWeight of the boulder is 71 x 9.8 N = Fwill + Fcliff\n\nFor the balancing of turning effect condition:\n(By the principle of moments) The distances Dcliff and Dwill from the CM of the two forces should be in inverse proportion  of the two forces .: \n\nDcliff = 3.15m and  \nDwill = (3.50-3.15)\nleading to: \nFcliff  = (71 x 9.8) 3.15/3.50 =. 626.22 N and\nFwill  =  (71 x 9.8) (3.50 - 3.15 ) /3.50 =.69.58 N\n\nAnswer:\nCliif applies 626.22N and Will applies 69.58 N..Got it?
93876
A quick guess without having reviewed your math, try adding g = 9.8 m/sec^2 to your answer and see if that is the answer that is needed.
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
97560
Are you referring to a fluid mechanics problem, or?\n\nCan you restate the question?
533021
there are some solved physics problems on \nhttp://www.dctech.com/physics/help/\nbut I don't know if it will help you out.
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
153350
This problem employs the simple concept of conservation of energy (athough there are a number of ways to solve it).\n\nBasically, the person in the wheel chair has an initial kinetic energy at the bottom of thr ramp,\nKE = 1/2 mv^2\nAt the top of the ramp, some of that kinetic energy has been transfered into Gravitational Potential Energy,\nPE = mgh\n\nSince the mass of the person is constant, that drops out of the equation.\n\nSolve for the change in PE and subtract this from the intial KE.  Using the new KE, solve for the final velocity.
194029
Hi!\nI change my field of study from physics to biophysics but still its a great pleasure to solve a problem in physics if you dont mind.\nWish to see your project.\nBye
630060
Speed of.. gravity? :S
10103
Studying for a Physics exam. But got really bored and so decided to come answer a few questions!
976575
Hi Mark, I will answer your question if you answer mine first.\nWhy are you asking this question, it seems like a lot of people in school are trying to take the easy way out of a physics or math problem.\nWill answer if you are not doing said above!\nCheers, Susie
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
50475
A)come on, the weight of the man is just his mass times gravity.\nB)this sounds like a torque problem, you need to know where on the beam the man is standing... since the beam is in equilibrium, you know that the sum of all torque is 0. 9im going to use T for torque because i cant get the greek letter i usually use) T=Fr (force times the distance from the pivit point) so all torques directed up minus all torques directed down equal 0. so that means all torques pointing up equal all torques pointing down. ok, so that means that:\n Tbeam + Tman = Tcable.\n and the cable is at an angle so it would actually be FrsinX instead of just Fr.  so now since the beam is uniform, we say that it's weight is in the center, so r is half of the length of the beam. so\n Tbeam=Fr=(288)(2.6) and now we convert the mass of the man to newtons, just multiplying by gravity(which im rounding to 10 because i dont feel like getting a calculator) and getting 750 Newtons. so that means\n Tman=Fr=750r. this has to stay as r because you didnt tell me where on the beam the man is standing. it probably indicate's where he's standing on your diagram. r would be the distance between the wall and the man because we're making the wall our pivot point, because if the end with the cable was our pivot point, everything there(including the cable) would be 0 because it would be 0m away from the pivot point, so we couldnt solve for Ft... so you'll have to plug in the "r" for the man yourself.  and Tcable=FrsinX=F(5.2)sin67 where F is the force of tension which is what you are solving for overall, r is 5.2 because the cable is 5.2m (the length of the beam) away from the pivot point. so now, we plug into what i told you before(Tbeam + Tman = Tcable.) which is true because the beam is in equilibrium. so \n\nTbeam + Tman = Tcable\nFr    + Fr   = FrsinX\n(288)(2.6) + (750)(r) = F(5.20)sin67\n\nplugging in the r which should be on your diagram, you can solve for F whichi is the tension in the cable.
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
620382
Horizontal momentum is conserved throughout the problem. i.e., when half the mass fell downwards, all of the horizontal momentum must be carried in the second.\nThere's too much missing information for me to go farther than that...
574012
I've alread answered two of your questiions in my last answer....\n\nIn this one, the guys above are correct - however, I'd try working it out for myself if I was you. Your tutor will soon realise something is wrong if you are giving answers to a subject you don't understand. Read the chapter, ask you teacher if you need more help. Learn Newtonian mechanics for yourself.