Newton's Laws Applications Review

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Part C: Force-Mass-Acceleration Relationships

Use the approximation that g= ~10 m/s2 to fill in the blanks in the following diagrams.

13.

 Fgrav = m•g = ~800 N

∑Fy = may = (80 kg)•(2.0 m/s/s)

Fy = 160 N, down

The Fgrav (down) and the Fair (up) must add up to 160 N, down. Thus, Fair must be smaller than Fgrav by 160 N.

Fair = 640 N

14.

Since  Fgrav = m•g, m can be calculated to be ~70 kg (m=Fgrav/g).

Since ay = 0 m/s/s, Fnorm must equal Fgrav; so Fnorm = 700 N.

Fx = m•ax = (70 kg)•(5.0 m/s/s)

Fx = 350 N, left

(Note that the ·Fx direction is always the same as the ax direction.)

With Fapp being the only horizontal force, its value must be 350 N - the same as ·Fx.

15.

 

   Fgrav = m•g = ~800 N

Since there are two forces pulling upwards and since the sign is hanging symmetrically, each force must supply an upwards pull equal to one-half the object's weight. So the vert pull (Fy) in each force is 400 N. The following triangle can be set up:

Using trig, we can write:

sin(30 deg.)=(400 N)/Ftens

Solving for Ftens yields 800 N.



16.

 

A quick blank is Fgrav:  Fgrav = m•g = ~80 N

Now resolve the 60-N force into components using trigonometry and the given angle measure:

Fx = 60 N•cos(30 deg) = 52 N

Fy = 60 N•sin(30 deg) = 25 N

Since the acceleration is horizontal, the sum of the vertical forces must equal 0 N. So Fgrav = Fy + Fnorm.

Therefore Fnorm = Fgrav - Fy = 55 N.

Knowing Fnorm and mu, the Ffrict can be determined:

Ffrict = mu•Fnorm = 0.5*(55 N) = 27.5 N

Now the horizontal forces can be summed:

Fx = Fx + Ffrict = 52 N, right + 27.5 N, left

Fx = 24.5 N, right

Using Newton's second law, Fx = m•ax

So ax = (24.5 N)/(8 kg) = 3.1 m/s/s, right (3.06... m/s/s)

17.

 

The first step in an inclined plane problem is to resolve the weight vector into parallel and perpendicular components:

Fpar = m•g•sin(angle) = (420 N)•sin(30 deg) = 210 N

Fperp = m•g•cos(angle) = (420 N)•cos(30 deg) = 364 N

The mass can be found as m = Fgrav/g

m = Fgrav/g = (420 N)/(10 m/s/s) = ~42 kg

The Fnorm acts opposite of and balances the Fperp.

So Fnorm = Fperp = 364 N

Knowing Fnorm and mu, the Ffrict can be determined:

Ffrict = mu•Fnorm = 0.2*(364 N) = 73 N

Now the forces parallel to the incline can be summed:

F|| = F|| + Ffrict = 210, down to left + 73 N, up to right

F|| = 137 N, down to left

Using Newton's second law, F|| = m•a||

So a|| = (137 N)/(42 kg) = 3.3 m/s/s

18.

 

Treating the two masses as a single system, it can be concluded that the net force on the 9-kg system is:

Fsystem = m•asystem = (9 kg)•(2.5 m/s/s) = 22.5 N, right

The free-body diagram for the system is:

The Fnorm supporting the 9-kg system is ~90 N.

So the Ffrict acting upon the system is:

Ffrict = mu• Fnorm = 0.20*(90 N) = 18 N, left

So if Fsystem = 22.5 N, right and Ffrict = 18 N, left, the rightward Ftens1 must equal 40.5 N.

The Ftens2 force is found inside the system; as such it can not be determined through a system analysis. To determine the Ftens2, one of the individual masses must be isolated and a free-body analysis must be conducted for it.

The 3-kg mass is selected and analyzed:

The Fnorm and Fgrav balance each other; their value is ~30 N. The Ffrict on the 3-kg mass is:

Ffrict = mu•Fnorm = 0.20*(30 N) = 6 N, left

The net force on the 3-kg object is:

Fx = m•ax = (3 kg)•(2.5 m/s/s) = 7.5 N, right

The horizontal forces must sum up to the net force on the 3-kg object; So

Fx = m•ax = Ftens2 (right) + Ffrict (left)

7.5 N, right = Ftens2 + 6 N, left

The Ftens2 must be 13.5 N.

19.

 

 Like most two-body problems involving pulleys, it is usually easiest to forgo the system analysis and conduct separate free-body analyses on the individual masses. Free-body diagrams, the chosen axes systems, and associated information is shown below.

Analyzing the Fx forces on the 250-g mass yields:

max = Ftens - Ffrict

Since Ffrict = mu•Fnorm and Fnorm = 2.5 N

The Ffrict is (0.1)*(2.5 N) = 0.25 N.

Substituting into equation 1 yields

(0.250 kg)•ax = Ftens - 0.25 N

 

Analyzing the Fy forces on the 50-g mass yields:

may = Fgrav - Ftens

Substituting m and Fgrav values into equation 3 yeilds:

(0.050 kg)*ay = (0.500 N) - Ftens

The above equation can be rearranged to:

Ftens = (0.500 N) - (0.050 kg)*ay

Equation 4 provides an expression for Ftens; this can be substituted into equation 2:

(0.250 kg)•ax = (0.500 N) - (0.050 kg)*ay - 0.25 N

Now since both masses accelerate at the same rate, ax =ay

and the above equation can be simplified into an equation with 1 unknown - the acceleration (a):

(0.250 kg)•a = (0.500 N) - (0.050 kg)*a - 0.25 N

After a few algebra steps, the acceleration can be found:

(0.0300 kg)•a = 0.25 N

a = 0.833 m/s/s

Now that a has been found, its value can be substituted back into equation 4 in order to solve for Ftens:

Ftens = (0.500 N) - (0.050 kg)*(0.833 m/s/s)

Ftens = 0.458 N

20.

 

This problem can most easily be solved using separate free-body analyses on the individual masses. Free-body diagrams, the chosen axes systems, and associated information is shown below.

Note that the positive y-axis is chosen as being downards on the 200-g mass since that is the direction of its acceleration. Similarly, it chosen as upwards on the 100-g mass since that is the direction of its acceleration.

For the 200-gram mass, the sum of the vertical forces equals the mass times the acceleration:

 Fgrav - Ftens = m•ay

2.00 N - Ftens = (0.200 kg)•ay

The same type of analysis can be conducted for the 100-gram mass:

Ftens - Fgrav = m•ay

Ftens - 1.00 N = (0.100 kg)•ay

Equation 2 can be rearranged to obtain an expression for the tension force:

Ftens = (0.100 kg)•ay + 1.00 N

This expression for Ftens can be substituted into equation 1 in order to obtain a single equation with acceleration (ay) as the unknown. The ay value can be solved for.

2.00 N -[(0.100 kg)•ay + 1.00 N] = (0.200 kg)•ay

2.00 N - 1.00 N = (0.200 kg)•ay + (0.100 kg)•ay

1.00 N = (0.300 kg)•ay

ay = (1.00 N)/(0.300 kg) = 3.33 m/s/s

Now with ay known, its value can be substituted into equation 3 in order to determine the tension force:

Ftens = (0.100 kg)•ay + 1.00 N

Ftens = (0.100 kg)•(3.33 m/s/s) + 1.00 N

Ftens = 0.333 N + 1.00 N = 1.33 N

21.

 

Like #20, this problem can most easily be solved using separate free-body analyses on the individual masses. Free-body diagrams, the chosen axes systems, and associated information is shown below. Note that in chosing the axis system, it has been assumed that object 1 will accelerate up the hill and object 2 will accelerate downwards. If this ends up to be false, then the acceleration values will turn out to be negative values.

Object 1 is on an inclined plane. The usual circumstances apply; their is no acceleration along what has been designated as the y-axis.

Fnorm = Fperp = m•g•cos(theta) = 888.2 N

The parallel component of Fgrav is

F|| = m•g•sin(theta) = (100 kg)•(9.8 m/s2)•sin(25)

F|| = 414.2 N

The Ffrict value can be found from the Fnorm value:

Ffrict = mu•Fnorm = (0.35)•(888.2 N) = 310.9 N

The Fx = m•ax equation can now be written:

Fx = m•ax

Ftens - Ffrict - F|| = m•ax

Ftens - 310.9 N - 414.2 N = m•ax

(Note that the Ffrict and F|| forces are subtracted from Ftens since they are heading in the direction of the negative x-axis.)

The above process can be repeated for object 2. The Fy = m•ay equation can now be written:

Fy = m•ay

Fgrav - Ftens = m•ay

(980 N) - Ftens = m•ay

The separate free-body analyses have provided two equations with two unknowns; the task at hand is to use these two equations to solve for Ftens and a.

Equation 2 can be re-written as

(980 N) - m•ay = Ftens

Since both objects accelerate together at the same rate, the ax for object 1 is equal to the ay value for object 2. The subscripts x and y can be dropped and a can be inserted into each equation.

(980 N) - m•a = Ftens

Equation 3 provides an expression for Ftens in terms of a. This expression is inserted into equation 1 in order to solve for acceleration. The steps are shown below.

(980 N) - m•a - 310.9 N - 414.2 N = m•a

= 2•m•a

254.9 N = 2•(100 kg)•a

1.27 m/s2 = a

The value of a can be re-inserted into equation 3 in order to solve for Ftens:

Ftens = (980 N) - m•a =(980 N) - (100 kg)•(1.27 m/s2)

Ftens = 853 N

 

Useful Web Links
Finding Acceleration || Net Force Problems Revisited || Inclined Planes


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