Which of the following forces must have a magnitude equal to mg

What is the average force that is exerted on the ball during launch? The x- components of the T 1 and T 2 and equivalent but in opposite directions. Sam is hover boarding down they hall at a constant velocity. Assuming there is friction, which of the following must be true? A man standing on a scale in an elevator notices that the scale reads 40 newtons greater than his normal weight.

Which type of movement of the elevator could cause this greater-than-normal reading? A man weighing Newtons is standing in an elevator. If the elevator rises with an acceleration of 0. A car travels up a hill at constant speed. Which of the following diagrams best represents the forces acting on the car at this instant?

How to Calculate the Magnitude of a Force in Physics

Three forces of equal magnitude act on the box shown. One force is upward, one downward, and one rightward. Which of the following statements must be true based on this information?

which of the following forces must have a magnitude equal to mg

Select two answers. Played times. Print Share Edit Delete. Live Game Live. Finish Editing. This quiz is incomplete! To play this quiz, please finish editing it. Delete Quiz.Hosted By. If we say that an object is under the influence of forces which are in equilibrium, we mean that the object is not accelerating. The following rules help to solve problems in which a body is acted on by three forces.

The sum of all the forces acting vertically upwards must have the same magnitude as the sum of all the forces acting vertically downwards. The sum of all the forces acting horizontally to the right must have the same magnitude as the sum of all the forces acting horizontally to the left.

The principle of moments : the sum of all the clock-wise moments about any point must have the same magnitude as the sum of all the anti-clockwise moments about the same point.

If the three forces are represented by arrows, drawn to scale and at the correct angles to each other with the arrows drawn "head to tail"they will form a closed triangle. In practice, when using condition 1we will often be considering the vertical and horizontal components of forces. If you can find a point about which all the turning effects are zero, then there can be no net moment and the body will not have any tendency to rotate.

In condition 4 we are using the triangle of forces idea: any one of the forces must be equal in magnitude but opposite in sense to the vector sum of the other two forces. The top point of the ladder is 3 m above the ground and the other end is 2.

The mass of the ladder is 2 0 kg. Find the magnitude and direction of the force exerted by the ground on the ladder and the magnitude of the force exerted by the wall on the ladder. The wall is described as smooth so it exerts no force of friction.

The ladder is described as uniform so the force of gravity can be considered to act at its mid point and, of course, acts vertically downwards. If we extend the lines of action of these two forces they meet at a point. Rule 3 above says that the line of action of the third force must also pass through this point. We will first use the fact that the forces acting on the ladder are in equilibrium vertically.

The two forces with vertical effects are, mg and the vertical component of F. These two forces must therefore have equal magnitudes, rule 1, above. The force R has a clock-wise turning effect about this point and the weight, mg, has an anti-clock-wise turning effect about the same point.Any force or combination of forces can cause a centripetal or radial acceleration.

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. You may use whichever expression for centripetal force is more convenient. Centripetal force F c is always perpendicular to the path and pointing to the center of curvature, because a c is perpendicular to the velocity and pointing to the center of curvature.

This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

Figure 1. The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the F cthe smaller the radius of curvature r and the sharper the curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case.

Thus the centripetal force in this situation is. Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for F c from the equation. The coefficient of friction found in Part 2 is much smaller than is typically found between tires and roads.

Force, Mass & Acceleration: Newton's Second Law of Motion

The car will still negotiate the curve if the coefficient is greater than 0. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn.

Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.

Figure 2. This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road.

A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

Let us now consider banked curveswhere the slope of the road helps you negotiate the curve. See Figure 3. Race tracks for bikes as well as cars, for example, often have steeply banked curves. For ideal bankingthe net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively.

In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.

which of the following forces must have a magnitude equal to mg

The only two external forces acting on the car are its weight w and the normal force of the road N. A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is.

Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless.

Figure 3. The car on this banked curve is moving away and turning to the left. Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked.Calculating magnitudes for forces is an important part of physics. Learning to do this with a single force and with the resultant force from two or more individual forces is an important skill for any budding physicist or anyone working on classical physics problems for school.

Find the resultant force from two vectors by first adding the x -components and y -components to find the resultant vector and then use the same formula for its magnitude. The first step to understanding what it means to calculate the magnitude of a force in physics is to learn what a vector is.

When you read a temperature of 50 degrees F, it tells you everything you need to know about the temperature of the object. A vector is different because it has a direction as well as a magnitude. If you watch a weather report, you'll learn just how fast the wind is traveling and in what direction.

This is a vector because it gives you that extra bit of information. So if something is traveling 10 miles per hour toward the northeast, the speed 10 miles per hour is the magnitude, northeast is the direction, and both parts together make up the vector velocity. In physics problems, east and north are usually replaced with x and y coordinates, respectively.

Think of the x coordinate of the force as the base of a triangle, the y component as the height of the triangle, and the hypotenuse as the resultant force from both components.

Extending the link, the angle the hypotenuse makes with the base is the direction of the force. Using x for the x -coordinate, y for the y -coordinate and F for the magnitude of the force, this can be expressed as:. In words, the resultant force is the square root of x 2 plus y 2. Using the example above:. For three-component forces, you add the z component to the same formula. You can work out the direction using trigonometry.

Force, Mass & Acceleration: Newton's Second Law of Motion

The identity best-suited to the task for most problems is:. This means you can use the components of the force to work it out. You can use the magnitude and the definition of either cos or sin if you prefer.

The direction is given by:. If you have two or more forces, work out the resultant force magnitude by first finding the resultant vector and then applying the same approach as above.

The only extra skill you need is finding the resultant vector, and this is fairly straightforward. The trick is that you add the corresponding x and y components together.Isaac Newton's First Law of Motion states, "A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force. That situation is described by Newton's Second Law of Motion. For a constant mass, force equals mass times acceleration. F is force, m is mass and a is acceleration.

The math behind this is quite simple. If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half. Newton expanded upon the earlier work of Galileo Galileiwho developed the first accurate laws of motion for masses, according to Greg Bothun, a physics professor at the University of Oregon.

Galileo's experiments showed that all bodies accelerate at the same rate regardless of size or mass.

Easy physics question help?

Newton also critiqued and expanded on the work of Rene Descartes, who also published a set of laws of nature intwo years after Newton was born. Descartes' laws are very similar to Newton's first law of motion. Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.

In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force. However, if the object is already in motion, or if this situation is viewed from a moving inertial reference frame, that body might appear to speed up, slow down, or change direction depending on the direction of the force and the directions that the object and reference frame are moving relative to each other.

The bold letters F and a in the equation indicate that force and acceleration are vector quantities, which means they have both magnitude and direction. The force can be a single force or it can be the combination of more than one force.

It is rather difficult to imagine applying a constant force to a body for an indefinite length of time. In most cases, forces can only be applied for a limited time, producing what is called impulse. For a massive body moving in an inertial reference frame without any other forces such as friction acting on it, a certain impulse will cause a certain change in its velocity.

The body might speed up, slow down or change direction, after which, the body will continue moving at a new constant velocity unless, of course, the impulse causes the body to stop.

There is one situation, however, in which we do encounter a constant force — the force due to gravitational acceleration, which causes massive bodies to exert a downward force on the Earth. Notice that in this case, F and g are not conventionally written as vectors, because they are always pointing in the same direction, down. The product of mass times gravitational acceleration, mgis known as weightwhich is just another kind of force.

Without gravity, a massive body has no weight, and without a massive body, gravity cannot produce a force. In order to overcome gravity and lift a massive body, you must produce an upward force m a that is greater than the downward gravitational force mg. If the rocket needs to slow down, speed up, or change direction, a force is used to give it a push, typically coming from the engine.

The amount of the force and the location where it is providing the push can change either or both the speed the magnitude part of acceleration and direction. Now that we know how a massive body in an inertial reference frame behaves when it subjected to an outside force, such as how the engines creating the push maneuver the rocket, what happens to the body that is exerting that force?

Live Science. Please deactivate your ad blocker in order to see our subscription offer.In the event of a set starting but not being completed then all bets will be void unless settlement of bets is already determined. If no next set played then bets on that set will be void.

The nominated set must be completed for bets to stand. Set Betting will be settled as 2-1 to the winner of the Match tie-break, and 3rd set winner will also be settled accordingly. Any bets taken on Next Game Winner or Next Game Score will be void if the next game turns out to be a Match tie-break, though any bets on Next Game First Point will stand.

Any bets taken in error for the Correct Score or Number of Games in the 3rd set will be void. See also the general rules on Total Games markets. Bets are offered for a player to win the nominated point. In the event of forfeited points, these will count for final settlement. If a tie-break is not played in the nominated set, all bets on these markets are void. All bets stand irrespective of whether or not the tie-break includes the awarding of a penalty point.

If the umpire awards the tie-break as a penalty game prior to the game starting, all bets on the tie-break will be void. If the tie-break is awarded as a penalty game whilst in progress, bets on tie-break winner will stand, but bets on tie-break score will be void (unless the only way the tie-break could be won was Any Other.

In the event of the tie-break not being completed through disqualification or retirement, all bets on the tie-break will be void, with the exception of tie-break total points as detailed above.

If the official outcome of a tie-break is unspecified (e. In the event of disqualification or retirement, bets will be void if the player has not yet been broken (unless there is no conceivable opportunity for them to serve again - in which case bets will be settled on the player having not been broken).

If the wrong player is indicated as (Svr), then any bets taken on Current or Next Game, Current or Next Game Score, Point Betting or Next Game First Point will be void, regardless of the result. For competitions where two legged ties have a Golden Set to decide which team progresses (in the event of the tie being tied in matches won), then for settlement purposes the Golden Set does not count. To Qualify will be settled on the team progressing to the next round of the specified competition, and includes the outcome of a Golden Set if played.

An unplayed or postponed match will be treated as a non-runner for settling purposes unless it is played within 48 hours of the original start time. The following markets will be void if the match is not completed, unless the specific market outcome is already determined:For individual set markets, in the event of the set not being completed bets will be void, unless the specific market outcome is already determined.

For In-Play Point Betting, bets are offered for a team to win the nominated point. In the event of the point not being played, due to the game or set ending, all bets on that point will be void. In the event of referee enforced point deductions, official results will be used for settlement purposes, with the exception of Race to Markets and Point Betting that have already been determined.

All match markets will be settled on regulation time, unless stated otherwise. Regulation time this must be completed for bets to stand unless otherwise stated. In the event of a match starting but not being completed then bets will be void, unless the specific market outcome is already determined.

In-Play Markets - Bets are settled on regulation time, unless stated otherwise. If the conditions of a specific event are changed from those originally listed by the official governing body then bets will be void, unless settlement of the bet(s) is already determined.

which of the following forces must have a magnitude equal to mg

Bets on any participant who takes part in qualifying for a specified event but then fails to qualify for the main Round(s) will be classed as losers.

For settlement purposes the result is at the time of the podium presentation. Specifically for Nordic Combined betting, if the results of the provisional competition round are used for the start of the cross country race, all event bets are void.

If one or more participants fail to take part in a specified event then bets will be void. If an event comprises of just one round then official results from that round will be used for settlement purposes.In many stats classes I've taken other places, homework assignments can seem punitive for those who don't understand the material, but this homework seemed to further enhance my learning experience.

I would definitely take another class with Dr. Pardoe if I had the opportunity. The interaction with the lecturer was good, the book is great, the online book material on software is extremely helpful and the lecturer put a lot of effort into a synthesis of the books contents every week. This was by far the best course I took at statistics. Course administration was efficient. Overall, a pleasant learning experience which is at least as good as any face-to-face course I've attended.

Rasch Applications Part 2 is a wonderful follow-up to the first course. Both courses are well designed and build up over the weeks in a such a manner that they make learning new concepts easy to handleThis course has given me a good understanding of the basics in Rasch analysis. I am very happy that I took this course because I do not think it is possible to just read this kind of information without working with data to get a good understanding of these complex theories and methods.

The instructor Joris Mays was very effective, especially in the discussion board where his feedback was timely, clear and very thoroughThis course was extremely helpful in facilitating my understanding of programming in R fundamentals.

I feel confident moving forward onto the next section of R programmingDr. Murrell is a great instructor. His notes and book were great resources.

I really appreciated how how he took time to make sure we understood the concepts. He is definitely the best (along with Dr. Verzani from the R stats course) instructor that I've had in all the classes I have taken here at statistics. It forced me to do it and that is what I needed.

Tal Galilli was great!. He is very patient and willing to help no matter what our questions were. Overall,great course and I plan to take more. This will be very helpful for the type of data analysis work I doThis course is an excellent follow up to the R basic course. A lot of new and helpful material was added and contributed to a more advanced understanding of R programmingProfessor John Verzani is excellent.

I used to be thrilled to see his long explanation on the concepts. It was so helpful and makes us love the subject. If you understand the subject, you start loving the subject and this is exactly what I realized and professor's explanation made me start loving R.


Replies to “Which of the following forces must have a magnitude equal to mg”

Leave a Reply

Your email address will not be published. Required fields are marked *