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Unbalanced Forces – Advanced Problem Solving DIRECTIONS: Read the following sections (including the example problems) and then complete the problems. Hopefully, at this point in the year, we understand the difference between balanced and unbalanced forces.

• Balanced Forces – two or more forces acting on the same object canceling the effects of the others, resulting in a net force of zero (𝚺𝑭 = 𝟎  𝑵).

o Motion characterized by constant velocity

• Unbalanced Forces – one or more forces acting on the same object resulting in a net force that is NOT zero (𝚺𝑭 ≠ 𝟎  𝑵).

o Characterized by accelerated motion During Unit 5 we have discussed Newton’s Second Law and how it can be used to relate mass, acceleration, and net force. Mathematically, we have expressed the relationship in Newton’s Second Law as

𝐚 =  𝚺𝑭𝒎

where 𝐚 is the acceleration of the system, 𝒎 is the mass of the system, and 𝚺𝑭 is the net force. We have also rearranged the above equation to

𝚺𝑭 =𝒎 ∙ 𝒂 Using Newton’s Second Law, we can find the acceleration of an object/system if we know the mass of the object/system and the net force. In Unit 3, we described the motion of objects using position vs. time graphs, velocity vs. time graphs, acceleration vs. time graphs and the following equations:

∆𝒙 =  𝒗𝒊 ∙ ∆𝒕+  𝟏𝟐𝒂 ∆𝒕 𝟐

𝒂 =  𝒗𝒇 − 𝒗𝒊∆𝒕

𝒗𝒇𝟐 = 𝒗𝒊𝟐 + 𝟐𝒂 ∙ ∆𝒙

∆𝒙 =  𝒗𝒊 + 𝒗𝒇𝟐   ∙  ∆𝒕

Using Newton’s Second Law and problem solving techniques from Unit 3, we are now able to solve many new problems!

Example Problem A car is travelling along the highway at a speed of 30 m/s (about 70 miles/hour). The car has a mass of 1200 kg, and the coefficient of friction between the tires and the road is 0.62. The driver sees a tree fall on the road in the distance. The driver immediately slams on the breaks. How far will the car travel before it comes to a stop? How long will this take? a.) How far will the car travel before it comes to a stop? We should always start by listing what we know. 𝑣! = 30  𝑚 𝑠 𝑣! = 0  𝑚 𝑠 (a stopped car has zero velocity) 𝑚 = 1200  𝑘𝑔 𝜇! = 0.62 Then we should figure out what forces act on the car by drawing a free body diagram (force diagram). We know the car is slowing down, so the forces are unbalanced. There must be a net force slowing the car down. We need to find the net force before we can find the acceleration, so we should probably complete a force table. 𝐹! = 𝑚 ∙ 𝑔

𝐹! = 1200  𝑘𝑔 ∙ −9.8  𝑚 𝑠! 𝑭𝒈 = −𝟏𝟏𝟕𝟔𝟎  𝑵

We know the car is not accelerating in the vertical direction, so the net force in thevertical direction is 0 N. Therefore, the normal force (𝐹!)will be opposite to the force of gravity. 𝐹! = 𝜇! ∙ 𝐹! 𝐹! = (0.62) ∙ (11760  𝑁) 𝑭𝒇 = −𝟕𝟐𝟗𝟏  𝑵 (Friction is acting to the left, so it will be negative) We know the net force is the sum of all the forces; so if we add the forces in our columns, we find the net force is -7291 N to the left. Using Newton’s Second Law, we can find the acceleration.

𝑎 =Σ𝐹𝑚

𝑎 =(−7291  𝑁)1200  𝑘𝑔

𝒂 = −𝟔.𝟎𝟖𝒎 𝒔𝟐

Horizontal Direction

Vertical Direction

Fg 0 N −11760  𝑁 FN 0 N 11760  𝑁 Ff −7291  𝑁 0 N

NET FORCE −𝟕𝟐𝟗𝟏  𝑵 0 N

FN  by  ground  on  car  

Fg  by  earth  on  car    

Ff  by  ground  on  car    

With the acceleration, we can refer back to problem solving methods in Unit 3. Lets list the information we now know. 𝑣! = 30  𝑚 𝑠 𝑣! = 0  𝑚 𝑠 (a stopped car has zero velocity) 𝑎 = −6.08𝑚 𝑠! Δ𝑥 =? ? ? (this is what we are looking for) We are looking for the change in position and we know the initial velocity, final velocity, and acceleration. Therefore, we will choose that has those for variables.

𝑣!! = 𝑣!! + 2𝑎 ∙ ∆𝑥

(0)! = (30)! + 2(−6.08) ∙ ∆𝑥 With some mathematical manipulation,

−900 = (−12.16) ∙ ∆𝑥

−900 = (−12.16) ∙ ∆𝑥

∆𝒙 = 𝟕𝟒.𝟎𝟏  𝒎 So, the car will travel 74.01 m before stopping! That’s over three quarters of a football field! b.) How long will this take? We should always start by listing what we know. 𝑣! = 30  𝑚 𝑠 𝑣! = 0  𝑚 𝑠 (a stopped car has zero velocity) 𝑎 = −6.08𝑚 𝑠! ∆𝑥 = 74.01  𝑚 So we can choose any of our kinematics equations to solve for the time. I will choose to use,

𝑎 =  𝑣! − 𝑣!∆𝑡

∆𝑡 =  𝑣! − 𝑣!𝑎

∆𝑡 =  0− 30−6.08

∆𝒕 =  𝟒.𝟗𝟑  𝒔

We could also use graphs to solve the problem. Acceleration is the slope of a velocity vs. time graph. Because the acceleration is -6.08 m/s2, we can create a velocity time graph with a slope of -6.08 m/s/s.

We know the slope of the velocity vs. time graph tells us acceleration, but the area under the graph tells us the change in position. We see that the velocity reaches a stop (0 m/s) at approximately 4.9 s.

𝒕 ≈ 𝟒.𝟗  𝒔 The graph makes a triangle so the area under the graph would be

𝐴 =12 𝑏 ∙ ℎ

So, the change in position would be

∆𝑥 =12 4.9  𝑠 ∙ 30

𝑚𝑠

∆𝒙 = 𝟕𝟑.𝟓  𝒎

                 1                          2                          3                          4                        5        t(s)  

30  27  

24  21  

18  15  

12  9  

6  3  

 

Velocity  (m

/s)