[ #1 | #2 | #3 | #4 | #5 | #6 | #7 | #8 | #9 | #10 | #11 | #12 | #13 | #14 | #15 | #16 | #17 | #18 | #19 | #20 | #21 | #22 | #23 | #24 | #25 | #26 | #27 | #28 | #29 | #30 ]

a. A ball is dropped from rest from the top of a building. Assume negligible air resistance.

 

In the absence of F_air, the only force acting upon the ball is gravity. It is a projectile.

b. After being thrown, a football is moving upwards and rightwards towards the peak of its trajectory. Assume negligible air resistance.

 

In the absence of F_air, the only force acting upon the ball is gravity. It is a projectile. Note that an upwards moving object does not need an upwards force. Only an upwards accelerating object requires an upwards force.

c. After reaching a terminal velocity, a falling skydiver then opens up the parachute.

 

When the terminal velocity was reached, F_air = F_grav. Then the parachute was opened, making Fair even greater than before. This is represented by the larger arrow.

d. An air track glider is gliding to the right at constant velocity.

 

 There is no rightwards force. A rightwards force would only be required if there is a rightwards acceleration. If the glider is gliding; there is no mention of it being pushed or pulled (F_app) and if there is a constant velocity, there must be balanced forces.

e. A car is skidding to a stop while travling to the right.

 

 There is no rightwards force. A rightwards force would only be required if there is a rightwards acceleration. If the car is skidding (wheels are locked), friction acts in a direction opposite its motion to slow it down.

f. A downward moving elevator (held by a cable) slows down.

 

The cable supplies the tension force. Since the elevator is moving downwards and slowing down, there must be more upwards force than the downwards gravity force.

g. A 25.0-N force is applied at a 30-degree angle to a crate in order to accelerate it rightward across a rough, horizontal surface.

 

The applied force is upwards and rightwards. Its upward component contributes to the upwards F_norm to balance the force of gravity. (Note the relative size of F_norm.) 

h. A picture hangs summetrically by two wires oriented at angles to the vertical.

The force exerted by a wire is a tension force. With two wires, there would be two upwards-pulling tension forces. The down force is gravity.

i. A large crate slowly accelerated down a steep and rough inclined plane.

There are only three forces present. The F_par and F_perp are merely components of gravity; they are not separate forces. The normal force is perpendicular to the surface (drawn in blue).

Answer: DEFG

These (DEFG) statements are always true. Statements ABC might be true but are not always true.

Answer: ACEG

An object that is at equilibrium can never be accelerating; its acceleration MUST be 0 m/s/s. Thus, A is an answer; and because an object in free-fall (E) and an object moving in a circle (C) are also accelerating, they must be counted as answers as well. (NOTE: an object moving in a circle is changing its direction and as such has an acceleration.) If there is a net force (G), then by definition the object is not at equilibrium. Choices B, D, F and H could be true of an object at equilibrium (though none of them are always true).

Answer: 100 kg

Mass is the amount of matter present in the object and is independent of the weight of the object. The weight or force of gravity acting upon an object depends upon the mass and the acceleration of gravity (F_grav = m•g). Because g is different on different planets and locations in the universe, the force of gravity or weight of an object will not always be the same. If Big Bubba goes to the moon, he will weigh less due to the decreased acceleration of gravity. However, he will still look the same, still have the same amount of matter, and still have a mass of 100 kg. KNOW the distinction between mass and weight.
 

Answer: 37 kg (approx.)

The acceleration of gravity on the moon is 1/6-th the value of earth's; this explains why the force of gravity is 1/6-th that of earth's. The force of gravity (F_grav) on the moon is calculated using the equation F_grav = m•g where g =1.6 m/s/s (on the moon). If Little Billie's weight on the moon is 1/6-th of 360 N, then Little Billie weighs 60 N on the moon. Inserting 60 N into the equation with 1.6 m/s/s as the value of g, Little Billie's mass can be calculated to be approximately 37 N. (Alternatively, simply find Billie's mass on earth from his weight of 360 N and then conclude that this mass value is not going to change when he goes to the moon.)

Answer: FALSE

A rightward acceleration would require a rightward net force. But if an object is moving rightward and slowing down (a leftwards acceleration), then there is certainly not a rightward net force and possibly not even a rightward force at all. For instance, if a car is moving rightward and skidding to a stop (with wheels locked), then there is no rightward force upon the car. The only horizontal force is a leftward force of friction which serves to slow the car down. (If you're getting stuck on this question and ones like them, it might be time to read the page titled "The Big Misconceptsion"; use the link below.)

Answer: 0 N

If the velocity is constant, then the acceleration is 0 m/s/s and the net force is zero. A net force is only required in order to accelerate an object. Yes, this means that an object can be moving to the right and NOT have a rightward's net force. (If you're getting stuck on this question and ones like them, it might be time to read the page titled "The Big Misconceptsion"; use the link below.)

Answer: FALSE

In this instance, the normal force could be equal to the force of gravity. But all that we can conclusively know is that the all the vertical forces sum up to 0 N. If the object is upon an incline, then the normal force will not be equal to the force of gravity. Or if there is another force with an upward or downward component, then the normal force is not equal to the force of gravity.

a. increase

b. decrease

c. remain the same

Answer: B

The normal force is equal to the perpendicular component of the weight vector. So F_norm = F_perp = m•g•cos(angle). If the incline angle is increased, the cos(angle) decreases towards 0; thus, the normal force decreases as well. Of course, the extreme is when the angle is 90 degrees and there is no normal force.

Answer: 60-degree angle to the horizontal

The tension in the wire will have the greatest impact upon its tendency to break. As the wire becomes most vertically oriented, the horizontal component of the tension force is reduced and the tension becomes less. The moral of the story -- to support the weight of a picture, one only needs to pull upwards, not leftwards and rightwards. Of course, the best arrangement would be two wires pulling completely vertically.

Answer: D

A 50-N force at 30 degrees would have two components - 43 N, east and 25 N, north. These two forces would be equal to the 50-N force at 30 degrees N of E. They are calculated as:

F_east = (50 N)*cos(30 deg)

F_north = (50 N)*sin(30 deg) 

Answer: AC

Only A and C could have a constant speed. Both B and D would have an acceleration that is directed rightwards. Do not confuse acceleration, motion, and force. A rightwards acceleration requires that there be a rightwards net force, On the other hand, a rightwards motion could have no net force, a leftward net force or a rightwards net force; it depends upon whether the object is slowing down, speeding up or maintaining a constant speed. Only A and C show a balance of forces, so only A and C are moving with constant speed. If you're getting stuck on this question and ones like them, it might be time to read the page titled "The Big Misconception"; use the link below.

13.

 F_grav = m•g = ~800 N

∑F_y = ma_y = (80 kg)•(2.0 m/s/s)

∑F_y = 160 N, down

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

F_air = 640 N

14.

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

Since a_y = 0 m/s/s, F_norm must equal F_grav; so F_norm = 700 N.

∑F_x = m•a_x = (70 kg)•(5.0 m/s/s)

∑F_x = 350 N, left

(Note that the ·F_x direction is always the same as the a_x direction.)

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

15.

 

   F_grav = 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 (F_y) in each force is 400 N. The following triangle can be set up:

Using trig, we can write:

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

Solving for F_tens yields 800 N.

16.

 

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

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

F_x = 60 N•cos(30 deg) = 52 N

F_y = 60 N•sin(30 deg) = 30 N

Since the acceleration is horizontal, the sum of the vertical forces must equal 0 N. So F_grav = F_y + F_norm.

Therefore F_norm = F_grav - F_y = 50 N.

Knowing F_norm and mu, the F_frict can be determined:

F_frict = mu•F_norm = 0.5*(50 N) = 25 N

Now the horizontal forces can be summed:

∑F_x = F_x + F_frict = 52 N, right + 25 N, left

∑F_x = 27 N, right

Using Newton's second law, ∑F_x = m•a_x

So a_x = (27 N)/(8 kg) = 3.4 m/s/s, right (3375... m/s/s)

17.

 

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

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

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

The mass can be found as m = F_grav/g

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

The F_norm acts opposite of and balances the F_perp.

So F_norm = F_perp = 364 N

Knowing F_norm and mu, the F_frict can be determined:

F_frict = mu•F_norm = 0.2*(364 N) = 73 N

Now the forces parallel to the incline can be summed:

∑F_|| = F_|| + F_frict = 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:

∑F_system = m•a_system = (9 kg)•(2.5 m/s/s) = 22.5 N, right

The free-body diagram for the system is:

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

So the F_frict acting upon the system is:

F_frict = mu• F_norm = 0.20*(90 N) = 18 N, left

So if ∑F_system = 22.5 N, right and F_frict = 18 N, left, the rightward F_tens1 must equal 40.5 N.

The F_tens2 force is found inside the system; as such it can not be determined through a system analysis. To determine the F_tens2, 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 F_norm and F_grav balance each other; their value is ~30 N. The F_frict on the 3-kg mass is:

F_frict = mu•F_norm = 0.20*(30 N) = 6 N, left

The net force on the 3-kg object is:

∑F_x = m•a_x = (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

∑F_x = m•a_x = F_tens2 (right) + F_frict (left)

7.5 N, right = F_tens2 + 6 N, left

The F_tens2 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 F_x forces on the 250-g mass yields:

ma_x = F_tens - F_frict

Since F_frict = mu•F_norm and F_norm = 2.5 N

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

Substituting into equation 1 yields

(0.250 kg)•a_x = F_tens - 0.25 N

 

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

ma_y = F_grav - F_tens

Substituting m and F_grav values into equation 3 yeilds:

(0.050 kg)*a_y = (0.500 N) - F_tens

The above equation can be rearranged to:

F_tens = (0.500 N) - (0.050 kg)*a_y

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

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

Now since both masses accelerate at the same rate, a_x =a_y

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 F_tens:

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

F_tens = 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:

 F_grav - F_tens = m•a_y

2.00 N - F_tens = (0.200 kg)•a_y

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

F_tens - F_grav = m•a_y

F_tens - 1.00 N = (0.100 kg)•a_y

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

F_tens = (0.100 kg)•a_y + 1.00 N

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

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

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

1.00 N = (0.300 kg)•a_y

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

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

F_tens = (0.100 kg)•a_y + 1.00 N

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

F_tens = 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.

F_norm = F_perp = m•g•cos(theta) = 888.2 N

The parallel component of F_grav is

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

F_|| = 414.2 N

The F_frict value can be found from the F_norm value:

F_frict = mu•F_norm = (0.35)•(888.2 N) = 310.9 N

The ∑F_x = m•a_x equation can now be written:

∑F_x = m•a_x

F_tens - F_frict - F_|| = m•a_x

F_tens - 310.9 N - 414.2 N = m•a_x

(Note that the F_frict and F_|| forces are subtracted from F_tens since they are heading in the direction of the negative x-axis.)

The above process can be repeated for object 2. The ∑F_y = m•a_y equation can now be written:

∑F_y = m•a_y

F_grav - F_tens = m•a_y

(980 N) - F_tens = m•a_y

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 F_tens and a.

Equation 2 can be re-written as

(980 N) - m•a_y = F_tens

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 = F_tens

Equation 3 provides an expression for F_tens 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 F_tens:

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

F_tens = 853 N

Answer: 26.8 m

Like most problems, this problem begins with a free-body diagram (as shown at right). Note that there is no rightwards applied force (a common mistake). Note also that the force of friction is the only force responsible for the acceleration (deceleration) of the car. The F_frict value is the net force. So determining the acceleration involves finding F_grav (F_grav=m•g = 945•9.8 = 9261 N), F_norm (the same as F_grav= 9261 N), and F_frict (F_frict = mu•F_norm = 0.972 • 9261 N = 9002 N). With the ∑F_x = 9002 N, the acceleration can be calculated (a_x = ∑F_x/m = 9002 N/945 kg = 9.53 m/s/s.) This is a leftwards acceleration; a_x will be assigned the numerical value of -9.53 m/s/s in the next part of this problem.

Now in the kinematic part of this problem, the distance must be found using the known information (v_f = 0 m/s, v_o = 22.6 m/s and a = 9.53 m/s/s). Use the equation:

Answer: 0.289 N

The acceleration of the object can be found from the slope of the line on the v-t graph. The acceleration is:

The net force is found from m•a; so F_net = (2 kg)•(0.125 m/s/s) = 0.250 N.

The only force that can contribute to the net force is the applied force. The horizontal component of the applied force must therefore be equal to 0.250 N. A diagram depicting the relationship of the lengths of the side for this 30-60-90 triangle is shown at the right. The cosine function can be utilized to determine the magnitude of the tension force.

Solving for F_app yields 0.289 N.

Answer: 56.6 N

If the mass of the sign is 10.0 kg, then the weight of the sign is 98 N (F_grav = m•g). This downward force of gravity must be balanced by the upward component of the tension. Thus, each cable must pull upwards with 49.0 N of force. In addition to the upward pull by the cable, there is also a horizontal pull. The tension force is the combined effect of these two componens. To determine the tension, set up a 60-30-90 triangle with a vertical component of 4.09 N as shown at the right. Then, use the sine function to solve for the hypotenuse of the triangle. The work is shown below:

Answer: All numerical answers are rounded to two significant digits:  F_frict= ~1040 N; a = ~6.2 m/s/s; d = ~44 m; v_f= 23 m/s 

This is a multi-part inclined plane problem. Like all problems, it should begin with a free-body diagram. If this is where the difficulty lies for you, then take some time to review inclined planes. See help page. The friction force opposes the motion of the boat. If the + x direction is defined as down and along the incline (as shown at the right) then the net force can be computed by the expression F_||- F_frict. The parallel component of the weight vector (m•g•sin theta) is 10395 N. The F_frict force is computed by multiplying the coefficient of friction (0.10) by the normal force (F_norm). The normal force is balancing F_perp(m•g•cos theta) and is equal to it. So F_norm= 10395 N and F_frict= 1039.4 N. The net force (F_net= F_||- F_frict ) is 9356 N; and the acceleration is F_net/m or 6.24 m/s/s.

Once the Newton's laws analysis has been completed and the acceleration determined, the kinematics portion of the problem can be tackled. First, determine the distance along the incline from the height of the hill and the angle of incline. The height of 100 ft is equivalent to 31.25 m; the distance along the incline is 44.19 m [found from (31.25 m)/sin (45)]. The final velocity is found using the acceleration, distance and a kinematic equation:

Answer: µ_static = 0.315; µ_kinetic = 0.246

A free-body diagram and a force analysis will yield the equation

for both the static and the kinetic situation. That is, when the object is at rest or moving down the incline at constant speed, the force down and parallel to the incline is balanced by the force which is up and parallel to the incline. The expressions for F_frict and F_|| can be substituted into this equation to yield

Knowing that F_norm is equal in magnitude to F_perp, the expression for F_perp can be substituted into the equation for F_norm. The new equation becomes

The angles for both the static and the kinetic case are known and mu (coefficient of friction) is the unknown. So the equation can be re-arranged to

The mass (m) and the acceleration of gravity (g) cancel from both the numerator and the denominator leaving sin(theta)/cos(theta) on the right side. This can be simplified again to

µ = tan(theta)

Equation 1 can be used to find the coefficient of friction for both the static and kinetic situation. The answers are as follows:

Answer: a = 1.52 m/s/s ; F_tens = 0.83 N

The solution to this problem begins by drawing a free-body diagram for each object. Note that the + x-axis for the 0.25-kg object is drawn in the direction that the object accelerates; and the + y-axis for the 0.10-kg object is drawn in the direction which it accelerates.

Note that while friction acts upon the 0.25-kg object, it does not act upon the 0.10-kg object (since it is not being dragged across the surface). Newton's second law of motion (·F = m•a) can be applied to the motion of the 0.25-kg mass:

The expressioin for F_frict can be substituted into the equation.

F_tens - µ•F_norm = m•a_x

Since the vertical forces on the 0.250-kg object balance each other, it is known that F_grav = F_norm; so F_norm = m•g = (0.250 kg)•(9.8 m/s/s) = 2.45 N. Since µ = 0.183, µ•F_norm = 0.448 N. Equation 1 can now be re-written as

F_tens - 0.448 N = (0.250 kg)•a_x

Newton's second law of motion (∑F = m•a) can be applied to the motion of the 0.10o-kg mass:

The expression for F_grav (m•g) can be substituted into the equation.

The mass of the 0.100-kg object can be substituted into the equation to yield equation 3.

0.98 N - F_tens = (0.100 kg)•a_y

Equations 2 and 3 each include an acceleration term - a_x and a_y. These acceleration values are the same since the system accelerates together; thus, a_x = a_y = a. An inspection of these two equations show that there are now two equations and two unknowns - F_tens and a. These unknown values can be solved for in the customary manner.

First, equation 3 is rearranged to create an expression for F_tens:

0.98 N - (0.100 kg)•a = F_tens

This expression for F_tens is now substituted into equation 2.

F_tens - 0.448 N = (0.250 kg)•a

0.980 N - (0.100 kg)•a - 0.448 N = (0.250 kg)•a

This equation can now be solved for acceleration (a) as shown in the following steps.

Now that a is known, its value can be substituted into equation 4 in order to solve for F_tens.

Answer: a = 3.95 m/s/s; F_tens = 20.6 N

Like #21 and #28, 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 on the back side of the hill. If this ends up to be false, then the acceleration values will turn out to be negative values.

The 22.7-N object is on an inclined plane. The usual circumstances apply; there is no acceleration along what has been designated as the y-axis.

The parallel component of F_grav is

The F_frict value can be found from the F_norm value:

The ∑F_x = m•a_x equation can now be written:

(Note that the F_frict and F_|| forces are subtracted from F_tens since they are heading in the direction of the negative x-axis.)

The above process can be repeated for the 34.5-N object. The ∑F_y = m•a_y equation can now be written:

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 F_tens and a.

Equation 2 can be re-written as

Since both objects accelerate together at the same rate, the a_x for the 22.7-N object is equal to the a_y value for the 34.5-N object. The subscripts x and y can be dropped and a can be inserted into each equation.

(34.5 N) - (3.52 kg) •a = F_tens

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

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

Answer: t =  3.95 s 

This problem involves a blending of Newton's laws and kinematics.  Newton's laws will have to be used to determine the acceleration;  then kinematics will have to be used to determine the time to fall 1.50 m from rest. The solution to this problem begins by drawing a free-body diagram for each object. Note that the + x-axis for the 500.0-gram object is drawn in the direction that the object accelerates; and the + y-axis for the 10.0-gram object is drawn in the direction which it accelerates.

Note that while friction acts upon the 500.0-gram object, it does not act upon the 10.0-gram object (since it is not being dragged across the surface). Newton's second law of motion (∑F = m•a) can be applied to the motion of the 0.5000-kg mass:

Newton's second law of motion (∑F = m•a) can also be applied to the motion of the 0.0100-kg hanging mass:

The expression for F_grav (m•g) can be substituted into the equation.

The mass of the 0.0100-kg object can be substituted into the equation to yield equation 2.

0.098 N - F_tens = (0.0100 kg)•a_y

Equations 1 and 2 include an acceleration term - a_x and a_y. These acceleration values are the same since the system accelerates together; thus, a_x = a_y = a. An inspection of these two equations show that there are now two equations and two unknowns - F_tens and a. These unknown values can be solved for in the customary manner. First, equation 2 is rearranged so as to express F_tens in terms of a.  This yields equation 3.

F_tens = 0.098 N - (0.0100 kg)•a_y