How is normal force related to frictional force ?
Q. I am zoning out on a problem dealing with an object on an incline plane. Accordingly, the object on the incline has two forces acting on it, the normal force and the frictional force. I just don't get why the object would not be in motion when the frictional frorce is less than or equal to the normal force.
Asked by darkness - Wed Jan 21 00:02:54 2009 - - 3 Answers - 0 Comments
A. Normal force is essentially equal to the pressure that an object puts on the surface upon which it rests. More normal force would equal more friction (heavier objects are harder to move, right??) On an inclined plane, it's the force of gravity down the slope that opposes friction. Normal force will always and forever be perpendicular to the force of friction, so it cannot oppose.
Answered by Chad P - Wed Jan 21 00:10:46 2009
Q. I am zoning out on a problem dealing with an object on an incline plane. Accordingly, the object on the incline has two forces acting on it, the normal force and the frictional force. I just don't get why the object would not be in motion when the frictional frorce is less than or equal to the normal force.
Asked by darkness - Wed Jan 21 00:02:54 2009 - - 3 Answers - 0 Comments
A. Normal force is essentially equal to the pressure that an object puts on the surface upon which it rests. More normal force would equal more friction (heavier objects are harder to move, right??) On an inclined plane, it's the force of gravity down the slope that opposes friction. Normal force will always and forever be perpendicular to the force of friction, so it cannot oppose.
Answered by Chad P - Wed Jan 21 00:10:46 2009
What is the magnitude of the normal force acting on the block?
Q. A force of 70.0 N is exerted at an angle of 30.0 below the horizontal on a block of mass 8.00 kg that is resting on a table. What is the magnitude of the normal force acting on the block?
Asked by Amber - Thu Feb 18 21:36:17 2010 - - 1 Answers - 0 Comments
A. so due to gravity there is an 80*9.81 N force and then add the y component of the 70 Newton force which would be 70sin(30) and your done.
Answered by Chris W - Thu Feb 18 21:41:33 2010
Q. A force of 70.0 N is exerted at an angle of 30.0 below the horizontal on a block of mass 8.00 kg that is resting on a table. What is the magnitude of the normal force acting on the block?
Asked by Amber - Thu Feb 18 21:36:17 2010 - - 1 Answers - 0 Comments
A. so due to gravity there is an 80*9.81 N force and then add the y component of the 70 Newton force which would be 70sin(30) and your done.
Answered by Chris W - Thu Feb 18 21:41:33 2010
What is the normal force of a rock lying on the ground?
Q. What is the normal force supporting a 18-kg rock lying on the ground in the following situations: A. The rock lies on the ground motionless B. 25-kg kid stands on top of the rock. C. A kid attempts to lift the rock with a force of 35 N.
Asked by Liz Frey - Sun Nov 15 12:04:48 2009 - - 5 Answers - 0 Comments
A. A. If no other forced are acting on the rock then Sum of the forces in the y direction=0 Fn-mg=0 Fn=mg Fn= (18)(9.81) Fn=176.58N B. Fn-mg(rock)-mg(kid)=0 Fn=mg(rock)+mg(kid) Fn=(18)(9.81)+(25)(9.81) Fn=421.83N c. Fn+Fy1-mg(rock)-mg(kid)=0 Fn+35= 176.58+(25)(9.81) Fn=386.83N
Answered by Breagh W - Sun Nov 15 12:14:26 2009
Q. What is the normal force supporting a 18-kg rock lying on the ground in the following situations: A. The rock lies on the ground motionless B. 25-kg kid stands on top of the rock. C. A kid attempts to lift the rock with a force of 35 N.
Asked by Liz Frey - Sun Nov 15 12:04:48 2009 - - 5 Answers - 0 Comments
A. A. If no other forced are acting on the rock then Sum of the forces in the y direction=0 Fn-mg=0 Fn=mg Fn= (18)(9.81) Fn=176.58N B. Fn-mg(rock)-mg(kid)=0 Fn=mg(rock)+mg(kid) Fn=(18)(9.81)+(25)(9.81) Fn=421.83N c. Fn+Fy1-mg(rock)-mg(kid)=0 Fn+35= 176.58+(25)(9.81) Fn=386.83N
Answered by Breagh W - Sun Nov 15 12:14:26 2009
rank the normal force of the car at different points on the hill?
Q. Suppose a car is moving with a constant speed along a mountain road. At various times during its drive, the car's location will be (A) at the top of a hill (B) at a dip between two hills (C) along a level stretch of road Rank these locations from greatest to least based on the normal force between the car and the road. Ties are possible.
Asked by Taylor Mc - Mon Feb 8 16:05:46 2010 - - 1 Answers - 0 Comments
A. entering the dip will reduce the load force, hitting the bottom and turning back up the opposite side will increase the load force the other two cases are the same, except you might argue that gravity is infinitesimally less on a hill top
Answered by daSVgrouch - Mon Feb 8 16:28:30 2010
Q. Suppose a car is moving with a constant speed along a mountain road. At various times during its drive, the car's location will be (A) at the top of a hill (B) at a dip between two hills (C) along a level stretch of road Rank these locations from greatest to least based on the normal force between the car and the road. Ties are possible.
Asked by Taylor Mc - Mon Feb 8 16:05:46 2010 - - 1 Answers - 0 Comments
A. entering the dip will reduce the load force, hitting the bottom and turning back up the opposite side will increase the load force the other two cases are the same, except you might argue that gravity is infinitesimally less on a hill top
Answered by daSVgrouch - Mon Feb 8 16:28:30 2010
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
What is the acceleration of the penguin and what is the normal force it feels?
Q. A 30 kg penguin slides down the side of a glacier that has a constant slope of 50 degrees. What is the acceleration of the penguin and what is the normal force it feels?
Asked by juju - Wed Jun 9 16:11:19 2010 - - 1 Answers - 0 Comments
A. Assuming ue = 0, a = g*sin50 = 7.50 m/s Fn = m*g*cos = 189 N
Answered by Steve - Wed Jun 9 16:20:46 2010
Q. A 30 kg penguin slides down the side of a glacier that has a constant slope of 50 degrees. What is the acceleration of the penguin and what is the normal force it feels?
Asked by juju - Wed Jun 9 16:11:19 2010 - - 1 Answers - 0 Comments
A. Assuming ue = 0, a = g*sin50 = 7.50 m/s Fn = m*g*cos = 189 N
Answered by Steve - Wed Jun 9 16:20:46 2010
How large is the normal force of the slide on the child?
Q. A 23kg child goes down a straight slide inclined 38degree above horizontal. The child is acted on by his weight, the normal force from the slide, and kinetic friction. I need help...I don't know how to solve this problem... Thanks in advance.
Asked by coldplayer415 - Fri Sep 26 19:59:50 2008 - - 1 Answers - 0 Comments
A. Assuming the child in in equilibrium (no acceleration/ moving at constant velocity) the forces must have a net value of 0. R = 23gcos38 = 180N (helps to draw/imagine a freebody diagram to see this) also the friction: F =23gsin38 = 140N
Answered by jsk678 - Fri Sep 26 20:09:51 2008
Q. A 23kg child goes down a straight slide inclined 38degree above horizontal. The child is acted on by his weight, the normal force from the slide, and kinetic friction. I need help...I don't know how to solve this problem... Thanks in advance.
Asked by coldplayer415 - Fri Sep 26 19:59:50 2008 - - 1 Answers - 0 Comments
A. Assuming the child in in equilibrium (no acceleration/ moving at constant velocity) the forces must have a net value of 0. R = 23gcos38 = 180N (helps to draw/imagine a freebody diagram to see this) also the friction: F =23gsin38 = 140N
Answered by jsk678 - Fri Sep 26 20:09:51 2008
How do i find the magnitude of the normal force?
Q. A 61 kg skier speeds down a trail, as shown in Figure 5-24. The surface is smooth and inclined at an angle of = 21 with the horizontal. Determine the magnitude of the normal force acting on the skier.
Asked by Tre T - Wed Dec 9 19:51:47 2009 - - 1 Answers - 0 Comments
A. You need to determine what component of the skier's weight vector is perpendicular to the surface. Copy the triangle available on Figure 5-24, and draw your skier on it. Draw his weight vector (61 kg * 9.8 m/s^2) straight down. This vector is his total weight, but only a portion of that weight is perpendicular to the surface of the hill. This vector will be the hypotenuse of two component vectors; one perpendicular to the hill and one parallel with the hill. Rotate the paper 21.8 degrees (so that the skier looks like he is on a flat surface, not a hill). Make a right triangle, using the weight vector as the hypotenuse. The trigonometry is hard to explain with text, but the angle on this triangle will also magically be 21.8 degrees. … [cont.]
Answered by Joelogs - Wed Dec 9 20:11:59 2009
Q. A 61 kg skier speeds down a trail, as shown in Figure 5-24. The surface is smooth and inclined at an angle of = 21 with the horizontal. Determine the magnitude of the normal force acting on the skier.
Asked by Tre T - Wed Dec 9 19:51:47 2009 - - 1 Answers - 0 Comments
A. You need to determine what component of the skier's weight vector is perpendicular to the surface. Copy the triangle available on Figure 5-24, and draw your skier on it. Draw his weight vector (61 kg * 9.8 m/s^2) straight down. This vector is his total weight, but only a portion of that weight is perpendicular to the surface of the hill. This vector will be the hypotenuse of two component vectors; one perpendicular to the hill and one parallel with the hill. Rotate the paper 21.8 degrees (so that the skier looks like he is on a flat surface, not a hill). Make a right triangle, using the weight vector as the hypotenuse. The trigonometry is hard to explain with text, but the angle on this triangle will also magically be 21.8 degrees. … [cont.]
Answered by Joelogs - Wed Dec 9 20:11:59 2009
If gravity and normal force are equal & Pull force and friction are equal. How can a object have force on it?
Q. Newtons 3rd law states an an object has an opposite and equal reaction for an action. So a crate on the ground has no force on it since the normal force and gravity are equal. When you pull the crate, the friction is equal to the pull force. How can a object move is no force is on it?
Asked by hmm - Mon Oct 19 03:14:04 2009 - - 4 Answers - 0 Comments
A. If the friction is equal to the pull force , then the crate will not move. The crate will move only if the pull force is greater than the frictional force. This makes it seem that the 3rd Law is a nonsense but it isn't. You have to look at the connections to Earth of the pull force to properly add up all the forces involved, and it turns out that they have always been found to add up to zero. Newton was quite remarkable in realising this and was made fun of by a heap of folk who didn't understand what was meant by a "system of forces".
Answered by rex - Mon Oct 19 03:26:56 2009
Q. Newtons 3rd law states an an object has an opposite and equal reaction for an action. So a crate on the ground has no force on it since the normal force and gravity are equal. When you pull the crate, the friction is equal to the pull force. How can a object move is no force is on it?
Asked by hmm - Mon Oct 19 03:14:04 2009 - - 4 Answers - 0 Comments
A. If the friction is equal to the pull force , then the crate will not move. The crate will move only if the pull force is greater than the frictional force. This makes it seem that the 3rd Law is a nonsense but it isn't. You have to look at the connections to Earth of the pull force to properly add up all the forces involved, and it turns out that they have always been found to add up to zero. Newton was quite remarkable in realising this and was made fun of by a heap of folk who didn't understand what was meant by a "system of forces".
Answered by rex - Mon Oct 19 03:26:56 2009
What is the significance of the slope of a Frictional Force vs Normal force plot?
Q. WHat is the significance of this slope? With frictional force as the y-axis and normal force on the x-axis?
Asked by magicharry17 - Wed Jan 6 23:39:09 2010 - - 1 Answers - 0 Comments
A. The slope will give you the roughness coefficient of friction.
Answered by gintable - Wed Jan 6 23:45:59 2010
Q. WHat is the significance of this slope? With frictional force as the y-axis and normal force on the x-axis?
Asked by magicharry17 - Wed Jan 6 23:39:09 2010 - - 1 Answers - 0 Comments
A. The slope will give you the roughness coefficient of friction.
Answered by gintable - Wed Jan 6 23:45:59 2010
How do u find the magnitude of the normal force?
Q. A 3.3 kg bag of groceries is in equilibrium on an incline of angle = 22 . Find the magnitude of the normal force on the bag.
Asked by sweetlatina - Sun Oct 14 03:10:06 2007 - - 3 Answers - 0 Comments
A. m*g* cos 22 when the x axis is chosen to point up the plane (which is the common convention)(where m = mass and g = 9.8 m/s^2). m*g*sin22 would be the force of gravity pointed down the x axis of the plane.
Answered by BJ - Sun Oct 14 03:23:47 2007
Q. A 3.3 kg bag of groceries is in equilibrium on an incline of angle = 22 . Find the magnitude of the normal force on the bag.
Asked by sweetlatina - Sun Oct 14 03:10:06 2007 - - 3 Answers - 0 Comments
A. m*g* cos 22 when the x axis is chosen to point up the plane (which is the common convention)(where m = mass and g = 9.8 m/s^2). m*g*sin22 would be the force of gravity pointed down the x axis of the plane.
Answered by BJ - Sun Oct 14 03:23:47 2007
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 Grant - 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 Grant - 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
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
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
How do I calculate the normal force?
Q. A cup and saucer rest on a table top. The cup has mass 0.176 kg and the saucer 0.165 kg. Calculate the magnitude of the normal force (a) the saucer exerts on the cup and (b) the table exerts on the saucer.
Asked by sLo - Sat Sep 27 02:32:36 2008 - - 4 Answers - 0 Comments
A. Newton's third law states that: Every object has an opposite and equal reaction. Becuse the objects are not falling, their downward forces are equal and opposite to their normal reaction forces (eg 1.5N down and 1.5N up. To calculate the downward force due to gravity (also known as weight), we use the equation W = mg, where "m" = mass and "g" = gravitational field strength (9.8). (Some text books will round "g" up to 10) So, a) the normal force the saucer exerts on the cup is equal and opposite to the downward force the cup exerts on the saucer. b) the normal force the table exerts on the saucer is equal and opposite to the combined downward force of the saucer and the cup. The forces are measured in Newtons (N).
Answered by Mike - Sat Sep 27 02:50:18 2008
Q. A cup and saucer rest on a table top. The cup has mass 0.176 kg and the saucer 0.165 kg. Calculate the magnitude of the normal force (a) the saucer exerts on the cup and (b) the table exerts on the saucer.
Asked by sLo - Sat Sep 27 02:32:36 2008 - - 4 Answers - 0 Comments
A. Newton's third law states that: Every object has an opposite and equal reaction. Becuse the objects are not falling, their downward forces are equal and opposite to their normal reaction forces (eg 1.5N down and 1.5N up. To calculate the downward force due to gravity (also known as weight), we use the equation W = mg, where "m" = mass and "g" = gravitational field strength (9.8). (Some text books will round "g" up to 10) So, a) the normal force the saucer exerts on the cup is equal and opposite to the downward force the cup exerts on the saucer. b) the normal force the table exerts on the saucer is equal and opposite to the combined downward force of the saucer and the cup. The forces are measured in Newtons (N).
Answered by Mike - Sat Sep 27 02:50:18 2008
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 Rosen sind blau und Ich liebe dich - 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 Rosen sind blau und Ich liebe dich - Tue Nov 3 09:52:58 2009
When an object is on an inclined plane, the normal force exerted by the inclined plane on the object is?
Q. (a) less than, (b) equal to, (c) more than the weight of the object. Why? Also if a 10 kg object on a 30 degree inclined plane, what are the object's weight and the normal force exerted on the object by the inclined plane?
Asked by Pascal - Sun Sep 2 17:45:52 2007 - - 3 Answers - 0 Comments
A. less than, otherwise it would slide down the plane? also its weight remains the same, but if the gravitaionl pull (which is both downwards directly below and also down the slope angle) overcomes the friction between the object and the plane then it slide downhill. If it was a 10kg block of ice on an icy slpoe it would slide down the slope. if it was a 10kg brick on a grass slope it wouldnt move. but the forces acting on both are the same
Answered by fast eddie - Sun Sep 2 17:56:48 2007
Q. (a) less than, (b) equal to, (c) more than the weight of the object. Why? Also if a 10 kg object on a 30 degree inclined plane, what are the object's weight and the normal force exerted on the object by the inclined plane?
Asked by Pascal - Sun Sep 2 17:45:52 2007 - - 3 Answers - 0 Comments
A. less than, otherwise it would slide down the plane? also its weight remains the same, but if the gravitaionl pull (which is both downwards directly below and also down the slope angle) overcomes the friction between the object and the plane then it slide downhill. If it was a 10kg block of ice on an icy slpoe it would slide down the slope. if it was a 10kg brick on a grass slope it wouldnt move. but the forces acting on both are the same
Answered by fast eddie - Sun Sep 2 17:56:48 2007
What is the magnitude of normal force?
Q. A(n) 89 kg block is pushed along the ceiling with a constant applied force of 1400 N that acts at an angle of 61 with the horizontal, as in the figure. The block accelerates to the right at 6.2 m/s2. The acceleration of gravity is 9.8 m/s2 . What is the magnitude of the normal force the ceiling exerts on the block? Answer in units of N. What is the coefficient of kinetic friction between the block and the ceiling?
Asked by Naria - Tue Mar 2 09:22:52 2010 - - 1 Answers - 0 Comments
A. The normal force in this problem would be the vertical component of the force applied to the wall minus the weight of the block. Therefore 1400sin(61)-89(9.8)=352.2 67N is the normal force. To find u you must find the frictional force. The force that should be to the right is equal to 1400cos(61)=678.733 N. However, friction is taking place. The actual force is equal to (89)(6.2)= 551.8 N. Subtract these values. 678.733-551.8=126.933 N. This is friction. Now use the equation Friction=Nu where N is the normal force. Plug in. 126.933=(352.267)u. So .36=u. Hope this helps!
Answered by ABC - Tue Mar 2 09:38:53 2010
Q. A(n) 89 kg block is pushed along the ceiling with a constant applied force of 1400 N that acts at an angle of 61 with the horizontal, as in the figure. The block accelerates to the right at 6.2 m/s2. The acceleration of gravity is 9.8 m/s2 . What is the magnitude of the normal force the ceiling exerts on the block? Answer in units of N. What is the coefficient of kinetic friction between the block and the ceiling?
Asked by Naria - Tue Mar 2 09:22:52 2010 - - 1 Answers - 0 Comments
A. The normal force in this problem would be the vertical component of the force applied to the wall minus the weight of the block. Therefore 1400sin(61)-89(9.8)=352.2 67N is the normal force. To find u you must find the frictional force. The force that should be to the right is equal to 1400cos(61)=678.733 N. However, friction is taking place. The actual force is equal to (89)(6.2)= 551.8 N. Subtract these values. 678.733-551.8=126.933 N. This is friction. Now use the equation Friction=Nu where N is the normal force. Plug in. 126.933=(352.267)u. So .36=u. Hope this helps!
Answered by ABC - Tue Mar 2 09:38:53 2010
How do you find normal force?
Q. A 120 lb man stands in an elevator that moves upward and has a downward acceleration of g/4. What is the normal force of the floor on the man? I already know that the answer to this is 90 lb, I just don't know how this number was found. Can someone help me please?
Asked by Stephanie S - Wed Feb 18 12:37:57 2009 - - 1 Answers - 0 Comments
A. The elevator is moving up and coming to a stop. Two forces act on the person: N, the normal force acts up mg, the person's weight, acts down Let [up] be positive and [down] negative. Fnet = ma N - mg = ma N = m(a+g) N = m(-g/4 + g) N = 3mg/4
Answered by unknown - Fri Feb 20 12:41:26 2009
Q. A 120 lb man stands in an elevator that moves upward and has a downward acceleration of g/4. What is the normal force of the floor on the man? I already know that the answer to this is 90 lb, I just don't know how this number was found. Can someone help me please?
Asked by Stephanie S - Wed Feb 18 12:37:57 2009 - - 1 Answers - 0 Comments
A. The elevator is moving up and coming to a stop. Two forces act on the person: N, the normal force acts up mg, the person's weight, acts down Let [up] be positive and [down] negative. Fnet = ma N - mg = ma N = m(a+g) N = m(-g/4 + g) N = 3mg/4
Answered by unknown - Fri Feb 20 12:41:26 2009
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 unknown - 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 unknown - Wed Dec 9 20:09:26 2009
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