I need help with forces?
958110
Putting it simply:\nEffort force, is force that must be applied to do something,eg, to push a block.\nResistance force, is the force which restricts the movement,eg, the friction of the block with the ground.
891370
Get the unbalanced force from the first 2 #s in your ? :  2.8-2.6 = .2\n\nb)  F = ma---> m = F/a = 0.2/.11 = 1.818 kg
517586
Moment = Force * Perpendicular Distance
1022160
I hope this will help. \nFor dynamics see reference 2
771573
OK, I have found both these forces.\n\nWhat should I do now?
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
746185
I can only answer 1 and 3 for you, cause i dont know the others, but here it goes:\n\n             1.) Think of this problem in terms of a right triangle. Draw a horizontal line. This line is the tidal force of 3* (10^3) N. Then draw a line on the end of the line perpendicular to the horizontal one. This line is your wind force of 6* (10^3) N. Connect the two ends of the lines with another line, to make a complete right angle triangle. This is where it gets a little tricky. The angle formed when the slanted line and the horizontal line meet is the angle you are trying to find. To find it, use the tan function, which is defined as opposite side's value (6*10^3 N) over the adjacent side's value (3*10^3 N). Lets use x as a variable for the unknown angle. So...\n\ntanx=(6*10^3) / (3*10^3)\nUse inverse of tan now (on calculators, you usually have to press shift or 2nd, then the tan key.) So...\n\nInverse tan ( (6*10^3) / (3*10^3) ) = x\n\ndo that in your calculator, and the value you get is the angle.\n\nFor the net force just subtract (3* 10^3) from (6*10^3) and the result will be your net force.\n\n3.) I dont know if u know this or not, but there is a formula that is part of Newton's second law. It is F=m*a.\n"F" being the Force, while "m" is the mass, and "a" is the acceleration. Just plug in the values you are given so that it comes out....\n\n390=270*a\n\n390/270=a ("a" is in meters per second squared)\n\nHope that helps
80396
Yeah i know:\nDo this:\n\na.)Find the work done by the applied force.\n\nb.)Find the work done by friction\n\nc.)Find the work done by gravity\n\nd.)Find the work done by the normal force.\n\ne.) IF the object starts at rest, how fast is it going after the 5.0m.\n\nhope this helped\n-----------\n\nNO I WILL NOT HLEP YOU. DO YOUR OWN WORK! your asking someone to do your work for you.
524636
If the forces are both acting in the same direction, add them together. = 65 Newtons
903133
From what I learned in physics class, the force you are speaking of is measured in Newtons.
696750
try the following. they are really helpful.\n\nwww.tutor4physics.com\nwww.physicsclassroom.com
611013
Oh good grief.  If it's anything like what I've just done it'll include velocity, centrepetal forces, magnetic fields, friction, gravity, forces generally, working out the mass of something and the speed at which it will fall, balances with pivot points, static electricity, AC & DC electricity, transformers, vectors and so on and so on.  It's quite hard but it's maths based so if you like maths, you should be ok.  I wish you luck.
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
857685
forces on M : grav. force Mg (vertically down) tension 2T(two strings horizontal, one on pulley) tension T (vertically down due to string on pulley) Normal reaction on M at lower surface N (vertically up) Normal reaction on M due to m R (to the left) Friction due to m on M f1=u1 times R (vertically down)(sliding is going on) friction in M at lower surface f2=u2 times N (to the left) (sliding is going on)\nnet force on M vertically up :ZERO\nto the right M times A\nforces on m  : grav force mg (vertically down) reaction due to M R (to the right) tension T (vertically up) friction due to M on m f1=u1 times R(vertically up)\nnet force on m to the right m times A\nnet force on m vertically down m times a\nunknowns N,R,a,A,T,f1,f2,\nwrite a constraint eqn for relating accns a and A. a=2A \nsolve them!!!!
710050
You have to ad the forces and show the resulting force. The equilibrium force will have the same magnitude and opposite direction (neutralizing all 5 forces)\n\nThink of the 1 force going North, the second one going East, the 3 going S, the 4 going W and the 5  NNE.\n\n2& 4 neutralize each other, so do not worry about them\n\n1 + 3 = 20 N 270 degrees\n\n5 = 25 N 0 deg + 43.3 N 90 deg  (50 cos 60 = 25, 50 sin 60 = 43.3)\n\nso the NS component will be 43.3 - 20 = 23.3 N 90 degrees\nand the EW will be 25N 0 deg\n\nThe resulting force will then be  34.17 =sqrt(23.3^2+25^2)\nwith direction 43 degrees (arctan[23.3/25])\n\nThe magnitude of the 6th force will then be 34.17 N and will have a direction of 223 degrees (opposite the 43 degrees one 43+180)\n\nyou can only arithmetically add forces that are in the same direction (or substract forces that have opposite direction).
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.