There are two situations to analyze in this Difficulty Level. Each involves the same set of knowns and unknowns. Numerical values will vary.

**Version 1:**

Analyze this: A 184-N force is applied at an angle of 20.4° above the horizontal to accelerate a 16.9-kg object across a level surface. The coefficient of friction is 0.156. Complete the diagram.

The two situations in this Difficulty Level are identical; only the numbers are different. There is a force applied at an angle to the horizontal. It's magnitude is stated; the angle (theta or θ) it makes with the horizontal is also stated. The coefficient of friction value is given. The mass is given. You must determine the x- and y-components, the force of gravity, the normal force, the friction force the net force and the acceleration. You can enter one value at a time and check its correctness or enter several values and check their correctness all at once. However you proceed, you will need to be mindful of the relationships between the various quantities. That's what this Help page is about. Our suggestions follow.

**First Things First**

There are four values provided; three of them fit with a blank. Read the **Analyze This** statement carefully, extract the numerical values and associate them with the quantities they represent. This is always the recommended starting point for any effective problem-solving strategy - identify what you know. Get that correct and you should have seven blanks left.

**Do the X-Y Thing**

The object accelerates horizontally across a surface. So you need to know all the horizontal force information. And it helps to also know all the vertical force information as well. But one of the forces is neither horizontal nor vertical (or is both horizontal and vertical). It's the force that makes this situation problematic - it's the angled applied force. But don't worry. Vector components are coming to the rescue.

Any force that makes an angle to the horizontal and vertical axis can be thought of as possessing a horizontal and a vertical part known as **components**. So your next task is to use the magnitude of the applied force (F_{app}) and the angle it makes with the horizontal (θ) to determine what the F_{x} and F_{y} values are. The diagram at the right should help. The x-component (F_{x}) and the y-component (F_{y}) form sides of a right triangle. The x-component is the side that is adjacent to the angle theta (θ); and so it is related to the hypotenuse (the applied force) by the cosine function. And since the y-component is the side opposite of the angle theta (θ), it is related to the hypotenuse by the sine function. And so we can assert that ...

**F _{x} = F_{app} • cosine θ**

F_{y} = F_{app} • sine θ

Perform the calculations, enter the values, and narrow down the missing blanks to four. And before you go on, **here's a TIP**: Now that you have resolved the angled force into two components, forget about the angled force. Disregard it. It's no longer important! It's called substitution. The applied force has been taken out of the ball game and replaced by two components. So don't worry about F_{app}. Do worry about F_{x} and F_{y}. The components are now the players in the game.

**Understand the Gravity of the Situation**

One of the blanks is the mass (**m**) and another blank is the weight or force of gravity (**F _{grav}**). And hopefully you know that they are related and you know the relationship. And if you don't, take some time to think about this relationship before you complete the analysis. It's a big thing in Physics. If it's confusing you now, it will confuse you later until you take the time to unconfuse yourself. The relationship is ...

**F _{grav} = m•g**

where

**g**= gravitational field strength (on Earth, approximately

**9.8 N/kg**)

The force of gravity is sometimes called the weight. It's value depends upon the mass of the object and the gravitational environment where the object is located. The **g** is called many things (some more accurate than others). But you should recognize that it reveals something about the gravitational environment where the object is located (probably, on Earth). We usually assume it refers to Earth where the value of g is 9.8 N/kg. Use this info to find one more blank ... and if you needed the equation to do this, then do yourself a favor and spend some time digesting this paragraph.

**Always Be Principled**

The normal force (**F _{norm}**) is the one purely upward force. This is the force of the supporting surface pushing up on the object. You might be conditioned to quickly equate it with F

_{grav}. That is, you might have something in your head that says ..."The normal force is always equal to the gravity force." THAT'S WRONG. It's often true that the normal force is equal to the gravity force but it is NOT ALWAYS TRUE (excuse the capitals here ... they're needed for needed emphasis). So read carefully:

The principle that leads to the faulty conclusion that "the normal force is always equal to the gravity force" is the principle that ...

**If there is no vertical acceleration, then the vertical forces are balanced.**

Student: So what's the problem? There's no vertical acceleration, right?

Teacher: Right!

Student: So vertical forces balance, right?

Teacher: Right!

Student: So F_{grav} and F_{norm} balance, right?

Teacher: Wrong!

Student: ???

If the vertical forces balance, then that means the two up forces (Yes. Two up forces.) must balance the one down force. There are two up forces! There's the normal force (which most students grow accustomed to) and there's the vertical component of the applied force - F_{y} (which may be new to many). And together, these two up forces equal the one down force such that ...

**F _{grav} = F_{norm} + F_{y}**

It's now time to substitute known values and calculate F_{norm}. You got this!

**What's That Fricken µ Symbol Mean?**

Excuse me but that fricken symbol is a friction symbol and it's called the coefficient of friction. It's a Greek letter; we Physics types call it "mu". (We even make cat and cow jokes about it ... but that's another story.) Mu or µ is not a force. It is a ratio of two forces. It's the ratio of the friction force to the normal force. It is specific to the two surfaces that slide across each other. It tells us how many Newtons of friction force that there are for every 1.00 N of force pressing the surfaces together (i.e., normal force). The useful equation that describes all this is ...

**F _{frict} = µ•F_{norm}**

You can (and should) use this equation to calculate the friction force. Do note though that the equation requires that you multiply the coefficient of friction (µ) by the normal force ... and not by the gravity force or any other force.

**Things Go Better With Newton**

Two blanks left - net force (**F _{net}**) and acceleration (

**a**). The net force is always the vector sum of all the individual forces; some teachers refer to it as ∑F. The vertical forces sum to zero; that is they're balanced (mentioned above). So focus on the horizontals. There's two horizontal - F

_{frict}and F

_{x}. The F

_{x}value is bigger; it is causing the acceleration. The F

_{frict}goes the opposite direction; it is opposing the acceleration. When you sum these two forces, you must call F

_{x}positive and F

_{frict}negative since they are directed in opposite directions. And so ...

**F _{net} = ∑F = F_{x} - F_{frict}**

Calculate the net force from the above. And once you know the net force, the acceleration can easily be determined using the Newton's second law equation ...

**a = F _{net} / m**

**Hey ... One More Thing ... and Perhaps the Most Important Thing**

If you relied on this Help page to do the first situation of the two situations in this Wizard Difficulty Level, then try this: Study your solution before you tap on the Check Answers button. Review all the logic. Then complete the second analysis without any reference to this page. That's how you learn! And that's how you know you learned!

Try one of these links to The Physics Classroom Tutorial for more help:

Net Force Problems Revisited