When drawing a free body diagram, is the upward normal force on a box from a table negative?
Q. So there is a box on a table. I get that the normal force is equal and opposite to the weight (but not an interaction pair). But since the weight is positive does that make the normal force negative? Or is everything just written as positive numbers?
Asked by Ana - Mon Jun 9 20:12:58 2008 - - 1 Answers - 0 Comments
A. Forces are vectors. When you have resolved forces into the forces that point up and down, you choose a direction to call positive and then the forces that point in that direction are positive and the forces that point in the opposite direction are negative. For a box sitting on a table, the vertical forces acting on it are its weight pointing down and the upward normal force exerted by the table. If you call up the positive direction then Fnormal + (-weight) = ma if the box is not accelerating, then a = 0, then Fnormal + (-weight) = m (0) or Fnormal = weight
Answered by Dennis H - Mon Jun 9 20:27:05 2008
Q. So there is a box on a table. I get that the normal force is equal and opposite to the weight (but not an interaction pair). But since the weight is positive does that make the normal force negative? Or is everything just written as positive numbers?
Asked by Ana - Mon Jun 9 20:12:58 2008 - - 1 Answers - 0 Comments
A. Forces are vectors. When you have resolved forces into the forces that point up and down, you choose a direction to call positive and then the forces that point in that direction are positive and the forces that point in the opposite direction are negative. For a box sitting on a table, the vertical forces acting on it are its weight pointing down and the upward normal force exerted by the table. If you call up the positive direction then Fnormal + (-weight) = ma if the box is not accelerating, then a = 0, then Fnormal + (-weight) = m (0) or Fnormal = weight
Answered by Dennis H - Mon Jun 9 20:27:05 2008
Calculate the normal force that the floor exerts on the chair?
Q. A chair of mass 8.8 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of 56.0 N that is directed at an angle of 30.4 degrees below the horizontal and the chair slides along the floor. Calculate the normal force that the floor exerts on the chair.
Asked by Colombia_Cali - Sat Mar 7 16:13:02 2009 - - 3 Answers - 0 Comments
A. When you push on the chair at 56 N, that push is both downward and horizontal...so by Newton's III Law (action = reaction, the floor is pushing back up an amount that not only equals the weight of the chair but the downward component of the force you are applying...so a little reflection shows that the total downward force on the floor must be F(dnwd) = 56 sin 30.4 + 8.8 X 9.8 = 114.6 N so the floor has to be pushing straight upward the same amount. Remember that W = mg so you have to multiply the MASS 8.8 kg by the acceleration of gravity 9.8 m/s^2 to get the WEIGHT.
Answered by radarsweep47 - Sat Mar 7 16:52:30 2009
Q. A chair of mass 8.8 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of 56.0 N that is directed at an angle of 30.4 degrees below the horizontal and the chair slides along the floor. Calculate the normal force that the floor exerts on the chair.
Asked by Colombia_Cali - Sat Mar 7 16:13:02 2009 - - 3 Answers - 0 Comments
A. When you push on the chair at 56 N, that push is both downward and horizontal...so by Newton's III Law (action = reaction, the floor is pushing back up an amount that not only equals the weight of the chair but the downward component of the force you are applying...so a little reflection shows that the total downward force on the floor must be F(dnwd) = 56 sin 30.4 + 8.8 X 9.8 = 114.6 N so the floor has to be pushing straight upward the same amount. Remember that W = mg so you have to multiply the MASS 8.8 kg by the acceleration of gravity 9.8 m/s^2 to get the WEIGHT.
Answered by radarsweep47 - Sat Mar 7 16:52:30 2009
What is the magnitude of the normal force on a block that is pushed with constant velocity?
Q. A block is pushed across a horizontal surface by the force F with constant velocity. F = 20N Angle = 30 degrees Mass = 10kg What is the magnitude of the normal force on the block?
Asked by Mon M - Mon Mar 2 22:09:08 2009 - - 1 Answers - 0 Comments
A. Fn = 20N*cos(30) + 10kg*9.81m/s^2
Answered by Chuck Norris - Mon Mar 2 22:15:21 2009
Q. A block is pushed across a horizontal surface by the force F with constant velocity. F = 20N Angle = 30 degrees Mass = 10kg What is the magnitude of the normal force on the block?
Asked by Mon M - Mon Mar 2 22:09:08 2009 - - 1 Answers - 0 Comments
A. Fn = 20N*cos(30) + 10kg*9.81m/s^2
Answered by Chuck Norris - Mon Mar 2 22:15:21 2009
When I stand up is the normal force causing the acceleration upwards?
Q. I am confused at how forces work. If I am stationary on hard ground there is force of gravity and the normal force. If I am squatting down and then I stand up I am accelerating as I start to stand up. I don't think Fgravity can change so how does the Fnormal change because of the acceleration? or is it the Fnormal that does not equal Fgravity anymore that is causing the acceleration? As you can see I am quite confused. All I know is that since there is movement there is now an acceleration butI dont really know how the forces work to cause the acceleration. thx! so why does standing up cuase a greater force downwards?
Asked by n33dh3lp - Fri Mar 9 01:23:55 2007 - - 4 Answers - 0 Comments
A. Ok, I'm kind of confused on how you asked this. I will explain the basic concepts of gravity and acceleration. Gravity for us on the scale that you're using is the Earth's pull on objects on the earth or around the earth (such as the moon). If you are stationary on hard ground, then there is a force of gravity acting down on your body. This causes you to remain on the ground and gives you weight. The real measurement using force would be the force of gravity times your mass. This would give you your weight in Newtons since force is measured in Newtons. You are pushing down on the Earth with your weight in newtons but there is also a force pushing back up on you, the Earth! If there was not this equal force pushing back on you, you would go… [cont.]
Answered by Ian B - Fri Mar 9 01:39:45 2007
Q. I am confused at how forces work. If I am stationary on hard ground there is force of gravity and the normal force. If I am squatting down and then I stand up I am accelerating as I start to stand up. I don't think Fgravity can change so how does the Fnormal change because of the acceleration? or is it the Fnormal that does not equal Fgravity anymore that is causing the acceleration? As you can see I am quite confused. All I know is that since there is movement there is now an acceleration butI dont really know how the forces work to cause the acceleration. thx! so why does standing up cuase a greater force downwards?
Asked by n33dh3lp - Fri Mar 9 01:23:55 2007 - - 4 Answers - 0 Comments
A. Ok, I'm kind of confused on how you asked this. I will explain the basic concepts of gravity and acceleration. Gravity for us on the scale that you're using is the Earth's pull on objects on the earth or around the earth (such as the moon). If you are stationary on hard ground, then there is a force of gravity acting down on your body. This causes you to remain on the ground and gives you weight. The real measurement using force would be the force of gravity times your mass. This would give you your weight in Newtons since force is measured in Newtons. You are pushing down on the Earth with your weight in newtons but there is also a force pushing back up on you, the Earth! If there was not this equal force pushing back on you, you would go… [cont.]
Answered by Ian B - Fri Mar 9 01:39:45 2007
what is the normal force that the spring exerts on the student?
Q. A spring scale is usually calibrated such that the mass that it indicates is related to the normal force that it exerted by the equation mscale = N/g , where N is the normal force exerted on the object that the scale is measuring the weight of, and g = 9.8 m/s2. If a student stand on this scale and the scale indicates that the student has a mass of 60 kg, what is the normal force that the spring exerts on the student?
Asked by Fggfg - Fri Oct 23 00:36:45 2009 - - 2 Answers - 0 Comments
A. mscale=N/g, so multiply both sides of the equation by g: mscale*g=(N/g)*(g/1) mscale*g=N Now we have: 60 kg*9.8 m/s^2=N 588 N=N, so the normal force is 588 Newtons.
Answered by stevemar2 - Mon Oct 26 18:40:44 2009
Q. A spring scale is usually calibrated such that the mass that it indicates is related to the normal force that it exerted by the equation mscale = N/g , where N is the normal force exerted on the object that the scale is measuring the weight of, and g = 9.8 m/s2. If a student stand on this scale and the scale indicates that the student has a mass of 60 kg, what is the normal force that the spring exerts on the student?
Asked by Fggfg - Fri Oct 23 00:36:45 2009 - - 2 Answers - 0 Comments
A. mscale=N/g, so multiply both sides of the equation by g: mscale*g=(N/g)*(g/1) mscale*g=N Now we have: 60 kg*9.8 m/s^2=N 588 N=N, so the normal force is 588 Newtons.
Answered by stevemar2 - Mon Oct 26 18:40:44 2009
Physics Help: Find the normal force that the ground applies to the front and rear tires?
Q. The mass of an automobile is 1200 kg, and the distance between its front and rear axles is 2.48 m. The center of gravity of the car is between the front and rear tires, and the horizontal distance between the center of gravity and the front axle is 1.02 m. (a) Determine the normal force that the ground applies to each of the two front wheels. (b) Determine the normal force that the ground applies to each of the two rear wheels.
Asked by lil_miz_britt06 - Wed Jun 25 16:50:28 2008 - - 2 Answers - 0 Comments
A. Because the center of gravity is closer to the front axle, the front wheels will experience a higher normal force from the ground than the rear wheels. Assuming only front/rear distances (not left/right), I'd set the problem up like this: Mass on Front (2.48m-1.02m)/2.48m (1200kg) = 0.57 * 1200kg = 684 kg Mass on Back (1.00-0.57) * 1200kg = 516 kg Force on each of the front wheels: F= ma = (684 kg * 9.8 m/s) / 2 wheels = 3351.6 N per wheel front. Force on each of the back wheels: F= ma = (516 kg * 9.8 m/s) / 2 wheels = 2528.4 N per wheel back. Hope that helps.
Answered by Phil - Wed Jun 25 18:20:59 2008
Q. The mass of an automobile is 1200 kg, and the distance between its front and rear axles is 2.48 m. The center of gravity of the car is between the front and rear tires, and the horizontal distance between the center of gravity and the front axle is 1.02 m. (a) Determine the normal force that the ground applies to each of the two front wheels. (b) Determine the normal force that the ground applies to each of the two rear wheels.
Asked by lil_miz_britt06 - Wed Jun 25 16:50:28 2008 - - 2 Answers - 0 Comments
A. Because the center of gravity is closer to the front axle, the front wheels will experience a higher normal force from the ground than the rear wheels. Assuming only front/rear distances (not left/right), I'd set the problem up like this: Mass on Front (2.48m-1.02m)/2.48m (1200kg) = 0.57 * 1200kg = 684 kg Mass on Back (1.00-0.57) * 1200kg = 516 kg Force on each of the front wheels: F= ma = (684 kg * 9.8 m/s) / 2 wheels = 3351.6 N per wheel front. Force on each of the back wheels: F= ma = (516 kg * 9.8 m/s) / 2 wheels = 2528.4 N per wheel back. Hope that helps.
Answered by Phil - Wed Jun 25 18:20:59 2008
How would i determine the magnitude of the normal force in this question?
Q. A 35-kg crate rests on a horizontal floor, and a 65-kg person is standing on the crate. Determine the magnitude of the normal force that: a) the floor exerts on the crate and b) the crate exerts on the person
Asked by angellica - Sat Oct 10 01:38:59 2009 - - 2 Answers - 0 Comments
A. First add together the total mass of the system, which equals 100kg. Multiply that by gravity, which is 9.80m/s^2. 980 N of force is exerted on the floor by the crate. In this case, since there are no angles involved and the system is in equilibrium, the normal force exerted on the crate by the floor must be 980N as well. For part B, do the same thing but just use the weight of the person on top of the crate. 65*9.8=637N and again, the normal force must be exactly the same.
Answered by Tony B - Sat Oct 10 01:47:34 2009
Q. A 35-kg crate rests on a horizontal floor, and a 65-kg person is standing on the crate. Determine the magnitude of the normal force that: a) the floor exerts on the crate and b) the crate exerts on the person
Asked by angellica - Sat Oct 10 01:38:59 2009 - - 2 Answers - 0 Comments
A. First add together the total mass of the system, which equals 100kg. Multiply that by gravity, which is 9.80m/s^2. 980 N of force is exerted on the floor by the crate. In this case, since there are no angles involved and the system is in equilibrium, the normal force exerted on the crate by the floor must be 980N as well. For part B, do the same thing but just use the weight of the person on top of the crate. 65*9.8=637N and again, the normal force must be exactly the same.
Answered by Tony B - Sat Oct 10 01:47:34 2009
What would be the normal force of the light fixture?
Q. You're holding up a light fixture of a mass 1.4 kg with a force of 21 N against the ceiling. What is the normal force? Thank You
Asked by Maxy - Wed Jul 8 16:45:25 2009 - - 1 Answers - 0 Comments
A. m*g - 21 if up is negative and g is 9.8 I get about 7 N
Answered by Old Science Guy - Wed Jul 8 16:56:41 2009
Q. You're holding up a light fixture of a mass 1.4 kg with a force of 21 N against the ceiling. What is the normal force? Thank You
Asked by Maxy - Wed Jul 8 16:45:25 2009 - - 1 Answers - 0 Comments
A. m*g - 21 if up is negative and g is 9.8 I get about 7 N
Answered by Old Science Guy - Wed Jul 8 16:56:41 2009
What is the minimum radius of the circle, so the normal force never exceeds three times his weight?
Q. A fighter pilot dives his plane toward the ground at 440 m/s. He pulls out of the dive on a vertical circle. What is the minimum radius of the circle, so that the normal force exerted on the pilot by his seat never exceeds three times his weight?
Asked by smallville427 - Sat Nov 10 11:24:23 2007 - - 1 Answers - 0 Comments
A. If R = normal reaction from the seat, R - mg = mv^2/r R = mg + mv^2/r < 3mg => v^2/r < 2g => r / v^2 > 1 / 2g => r > v^2 / 2mg = (40)^2 / 2 * 9.8 => r > 81.6 m Minimum radius = 82 m ( nearly)
Answered by Madhukar Daftary - Sat Nov 10 11:34:57 2007
Q. A fighter pilot dives his plane toward the ground at 440 m/s. He pulls out of the dive on a vertical circle. What is the minimum radius of the circle, so that the normal force exerted on the pilot by his seat never exceeds three times his weight?
Asked by smallville427 - Sat Nov 10 11:24:23 2007 - - 1 Answers - 0 Comments
A. If R = normal reaction from the seat, R - mg = mv^2/r R = mg + mv^2/r < 3mg => v^2/r < 2g => r / v^2 > 1 / 2g => r > v^2 / 2mg = (40)^2 / 2 * 9.8 => r > 81.6 m Minimum radius = 82 m ( nearly)
Answered by Madhukar Daftary - Sat Nov 10 11:34:57 2007
What does a constant normal force imply?
Q. In the case of an elevator going up with a constant normal force, does it imply that the acceleration of the elevator is constant? Also, in the same scenario above, would it be correct if I said that work = mass x g x height?
Asked by ArmedSquirrel - Thu Feb 19 20:16:44 2009 - - 1 Answers - 0 Comments
A. yes that means a is constant
Answered by mike - Thu Feb 19 20:30:38 2009
Q. In the case of an elevator going up with a constant normal force, does it imply that the acceleration of the elevator is constant? Also, in the same scenario above, would it be correct if I said that work = mass x g x height?
Asked by ArmedSquirrel - Thu Feb 19 20:16:44 2009 - - 1 Answers - 0 Comments
A. yes that means a is constant
Answered by mike - Thu Feb 19 20:30:38 2009
Why does the normal force of an object on a flat surface affect its friction?
Q. I know if you multiply mu times the normal force of an object as you drag it along the surface equals its friction, but I don't really understand why.
Asked by Ash - Sun Nov 22 20:21:52 2009 - - 2 Answers - 0 Comments
A. it has to do with the electrons of the object acting with the electrons of the surface
Answered by football6472 - Sun Nov 22 20:31:43 2009
Q. I know if you multiply mu times the normal force of an object as you drag it along the surface equals its friction, but I don't really understand why.
Asked by Ash - Sun Nov 22 20:21:52 2009 - - 2 Answers - 0 Comments
A. it has to do with the electrons of the object acting with the electrons of the surface
Answered by football6472 - Sun Nov 22 20:31:43 2009
Is the force of friction between two objects in contact directly proportional to the normal force?
Q. Is the force of friction between two objects in contact directly proportional to the normal force?
Asked by acddc - Tue Jan 5 20:28:22 2010 - - 1 Answers - 0 Comments
Q. Is the force of friction between two objects in contact directly proportional to the normal force?
Asked by acddc - Tue Jan 5 20:28:22 2010 - - 1 Answers - 0 Comments
How to find the normal force for an object at rest?
Q. Determine the normal force for a laundry basket with a mass of 4.9 kg in each of the following situation- at rest on a ramp inclined at 12 above the horizontal How do I do this?? I'm not asking for an answer, just how to figure it out
Asked by sincerelysarcastic - Wed Oct 28 21:51:26 2009 - - 1 Answers - 0 Comments
A. The forces acting perpendicular to the incline must be balanced. The perpendicular component of gravity is mgcos12 . The normal force must balance this component, therefore: n = mgcos12 .
Answered by David - Tue Nov 3 09:52:58 2009
Q. Determine the normal force for a laundry basket with a mass of 4.9 kg in each of the following situation- at rest on a ramp inclined at 12 above the horizontal How do I do this?? I'm not asking for an answer, just how to figure it out
Asked by sincerelysarcastic - Wed Oct 28 21:51:26 2009 - - 1 Answers - 0 Comments
A. The forces acting perpendicular to the incline must be balanced. The perpendicular component of gravity is mgcos12 . The normal force must balance this component, therefore: n = mgcos12 .
Answered by David - Tue Nov 3 09:52:58 2009
What is the min. r of the circle, so the normal force exerted never exceeds 3 times his weight?
Q. A fighter pilot dives his plane toward the ground at 440 m/s. He pulls out of the dive on a vertical circle. What is the minimum radius of the circle, so that the normal force exerted on the pilot by his seat never exceeds three times his weight?
Asked by Scuba Steve - Sat Nov 10 20:12:27 2007 - - 1 Answers - 0 Comments
A. To balance the gravitational pull on the driver, the seat must push him up with a force mg. To make him move along a circle of radius it must provide another force of mv^2 /r. But the total must not exceed 3mg. 2mg = mv^2 /r 2g = v^2 /r r = v^2 / 2g = 440^2 / 19.6 = 9877.6 5 m
Answered by Pearlsawme - Sat Nov 10 20:54:10 2007
Q. A fighter pilot dives his plane toward the ground at 440 m/s. He pulls out of the dive on a vertical circle. What is the minimum radius of the circle, so that the normal force exerted on the pilot by his seat never exceeds three times his weight?
Asked by Scuba Steve - Sat Nov 10 20:12:27 2007 - - 1 Answers - 0 Comments
A. To balance the gravitational pull on the driver, the seat must push him up with a force mg. To make him move along a circle of radius it must provide another force of mv^2 /r. But the total must not exceed 3mg. 2mg = mv^2 /r 2g = v^2 /r r = v^2 / 2g = 440^2 / 19.6 = 9877.6 5 m
Answered by Pearlsawme - Sat Nov 10 20:54:10 2007
How do you calculate Normal Force?
Q. I understand what normal force is, but my book doesn't explain how you would calculate it.
Asked by junipero locke - Sat Mar 24 17:59:14 2007 - - 5 Answers - 0 Comments
A. it's the force exerted by the surface something it laying. for example, if you have a box weighing 200N laying on the floor, and it's not accelerating, the normal force is equal to the force of gravity, 200N. but, if there is a rope pulling upwards on the box with a force of 100N and the box is not accelerating, the normal force is 100N. It's a simple manipulation of Newton's second law. F=ma. so, if a=0. then all opposing forces must be equal in magnitude (forces upwards = forces downward) In the second case, Fg=200N and the force from the rope (F1) = 100N. 200N = 100N + Fn (normal force). subtract 100 from both sides and you get Fn=100. That's normal force in the simplest context, if you would like more help email me at frickinaj@yahoo. [cont.]
Answered by frickinaj - Sat Mar 24 18:09:29 2007
Q. I understand what normal force is, but my book doesn't explain how you would calculate it.
Asked by junipero locke - Sat Mar 24 17:59:14 2007 - - 5 Answers - 0 Comments
A. it's the force exerted by the surface something it laying. for example, if you have a box weighing 200N laying on the floor, and it's not accelerating, the normal force is equal to the force of gravity, 200N. but, if there is a rope pulling upwards on the box with a force of 100N and the box is not accelerating, the normal force is 100N. It's a simple manipulation of Newton's second law. F=ma. so, if a=0. then all opposing forces must be equal in magnitude (forces upwards = forces downward) In the second case, Fg=200N and the force from the rope (F1) = 100N. 200N = 100N + Fn (normal force). subtract 100 from both sides and you get Fn=100. That's normal force in the simplest context, if you would like more help email me at frickinaj@yahoo. [cont.]
Answered by frickinaj - Sat Mar 24 18:09:29 2007
How do i find the normal force?
Q. Find the normal force exerted on a 2.2 kg book resting on a surface inclined at 23 above the horizontal.
Asked by Tre T - Wed Dec 9 19:56:43 2009 - - 1 Answers - 0 Comments
A. resolve the forces parallel and 90 degrees to the inclined plane. The normal force (at 90 degrees to the incline) is given by mass times gravity times cos angle Parallel by mass times gravity times sin angle So in your case 2.2 x 9.81 x cos 23 degrees =19.87 N (rounded) Hope this helps.
Answered by Stoorsooker - Wed Dec 9 20:09:26 2009
Q. Find the normal force exerted on a 2.2 kg book resting on a surface inclined at 23 above the horizontal.
Asked by Tre T - Wed Dec 9 19:56:43 2009 - - 1 Answers - 0 Comments
A. resolve the forces parallel and 90 degrees to the inclined plane. The normal force (at 90 degrees to the incline) is given by mass times gravity times cos angle Parallel by mass times gravity times sin angle So in your case 2.2 x 9.81 x cos 23 degrees =19.87 N (rounded) Hope this helps.
Answered by Stoorsooker - Wed Dec 9 20:09:26 2009
How does friction vary with the strength of the normal force?
Q. and how does varying the normal force affect the values of coefficient of friction of both static and kinetic friction?
Asked by ako - Sun Jan 24 21:05:00 2010 - - 1 Answers - 0 Comments
A.
Answered by Sara - Sun Jan 24 21:11:25 2010
Q. and how does varying the normal force affect the values of coefficient of friction of both static and kinetic friction?
Asked by ako - Sun Jan 24 21:05:00 2010 - - 1 Answers - 0 Comments
A.
Answered by Sara - Sun Jan 24 21:11:25 2010
What is the relationship between the force of friction and the Normal force?
Q. thank you so much this is a physics lab question dealing with To determine the forceful strength of static and kinetic friction and the effect they have with the other forces (gravity and normal Force).
Asked by Time2playthegame - Mon Oct 27 23:15:21 2008 - - 1 Answers - 0 Comments
A. force friction divided by force normal = a constant called the coefficient of friction '
Answered by hello - Mon Oct 27 23:23:28 2008
Q. thank you so much this is a physics lab question dealing with To determine the forceful strength of static and kinetic friction and the effect they have with the other forces (gravity and normal Force).
Asked by Time2playthegame - Mon Oct 27 23:15:21 2008 - - 1 Answers - 0 Comments
A. force friction divided by force normal = a constant called the coefficient of friction '
Answered by hello - Mon Oct 27 23:23:28 2008
Physics - How does the force of friction impact the normal force?
Q. F=ma Normal force = reaction to the weight of an object sitting on a surface - the same force this object exerts on the surface is rebounded back to the object
Asked by Aint No Bugs On Me - Tue Oct 20 17:49:56 2009 - - 1 Answers - 0 Comments
A. It is not quite right to say that the force the object exerts on the surface is rebounded back into the object. There are actually two distinct forces there: the force that the object exerts on the surface, and the force that the surface exerts on the object. They will be equal in magnitude and opposite in direction, by Newton's third law. Now, the force of friction acting on an object is typically proportional to the normal force acting on the object by the surface. The proportionality constant is the coefficient of friction, which produces the equation F = N where F is the frictional force, N is the Normal force, and is the coefficient of friction. In the case of kinetic friction, this will be the force acting opposite the object's [cont.]
Answered by Quantum Cop - Tue Oct 20 18:03:53 2009
Q. F=ma Normal force = reaction to the weight of an object sitting on a surface - the same force this object exerts on the surface is rebounded back to the object
Asked by Aint No Bugs On Me - Tue Oct 20 17:49:56 2009 - - 1 Answers - 0 Comments
A. It is not quite right to say that the force the object exerts on the surface is rebounded back into the object. There are actually two distinct forces there: the force that the object exerts on the surface, and the force that the surface exerts on the object. They will be equal in magnitude and opposite in direction, by Newton's third law. Now, the force of friction acting on an object is typically proportional to the normal force acting on the object by the surface. The proportionality constant is the coefficient of friction, which produces the equation F = N where F is the frictional force, N is the Normal force, and is the coefficient of friction. In the case of kinetic friction, this will be the force acting opposite the object's [cont.]
Answered by Quantum Cop - Tue Oct 20 18:03:53 2009
Physics- Is there a normal force that acts on dry ice and hovercrafts?
Q. Although normal force is usually a contact force, is there an upwards normal force acting on objects like hovercrafts and dry ice, which don't directly touch a surface, but are kept in the air by a small cushion of air underneath?
Asked by Maddie! - Sat Dec 5 19:45:44 2009 - - 2 Answers - 0 Comments
A. probably not youre right normal force is a contact force i didn't even know dry ice floats to be frank with you but hovercrafts float for the same reason rockets can ignite and blast off into space you have to remember newton's third law, there is an equal and opposite force for every force. the only reason you dont notice it is because equal forces doesnt necessarily imply equal distance moved that depends on the mass of each object that is why you need to burn a HUGE amount of fuel to launch a rocket and for that same reason, that is why hovercrafts must basically push the air beneath it with a huge amount of force so that the air molecules can give it an equal and opposite force that can move the hovercraft off of th land. i suppose… [cont.]
Answered by bpmonkey - Sat Dec 5 19:56:47 2009
Q. Although normal force is usually a contact force, is there an upwards normal force acting on objects like hovercrafts and dry ice, which don't directly touch a surface, but are kept in the air by a small cushion of air underneath?
Asked by Maddie! - Sat Dec 5 19:45:44 2009 - - 2 Answers - 0 Comments
A. probably not youre right normal force is a contact force i didn't even know dry ice floats to be frank with you but hovercrafts float for the same reason rockets can ignite and blast off into space you have to remember newton's third law, there is an equal and opposite force for every force. the only reason you dont notice it is because equal forces doesnt necessarily imply equal distance moved that depends on the mass of each object that is why you need to burn a HUGE amount of fuel to launch a rocket and for that same reason, that is why hovercrafts must basically push the air beneath it with a huge amount of force so that the air molecules can give it an equal and opposite force that can move the hovercraft off of th land. i suppose… [cont.]
Answered by bpmonkey - Sat Dec 5 19:56:47 2009
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