The Physics
Opus in profectus

Forces in Two Dimensions

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practice problem 1

A 4.5 kg Canada goose is about to take flight. It starts from rest on the ground, but after a single step it is completely airborne. After 2.0 s of horizontal flight the bird has reached a speed of 6.0 m/s (fast enough to stay aloft, but not so fast that we need to worry about air resistance… at first).
  1. Draw a free body diagram of the goose in flight.
  2. Determine the following quantities for the goose in flight…
    1. its acceleration
    2. its weight
    3. the magnitude and direction of the net force acting on it
    4. the magnitude of the upward lift provided by its wings
    5. the magnitude of the forward thrust provided by its wings
  3. Any object moving through the air will experience air resistance. We just decided to ignore it temporarily. If we now admit that air resistance was present to some extent, how will this change the computed values of…
    1. the acceleration?
    2. the weight?
    3. the net force?
    4. the lift?
    5. the thrust?
  • All the measurements given in the problem are still valid for part c of this problem. The mass is still 4.5 kg and the bird still accelerates from rest to 6.0 m/s in 2.0 s.


The solutions…

  1. All objects have weight. It points down. A winged object like a bird needs some force to keep it aloft. In aerodynamics, they call it lift. An object accelerating forward must have some force pushing it forward. That one's usually called thrust. If there was drag, it would point opposite the direction of bird's motion.

    1. In this problem, acceleration is computed from its definition.
      a =  v
      a =  6.0 m/s − 0 m/s
      2.0 s
      a = 3.0 m/s2 forward  
    2. Weight is mass times gravity (the value of the gravitational field on earth). Weight always points down. (Down is defined as the direction things move when allowed to fall freely.)
      W = mg
      W = (4.5 kg)(9.8 m/s2)
      W = 44.1 N down
    3. Use Newton's second law of motion. Net force and acceleration always have the same direction.
      F = ma
      F = (4.5 kg)(3.0 m/s2)
      F = 13.5 N forward
    4. There is no net force in the vertical direction. Thus, whatever forces act up must be balanced by forces acting down. According to our diagram, this means lift equals weight.
      L = −W = 44.1 N up
    5. The forward force of thrust is the only force acting horizontally, which makes it the net force.
      T = ∑F = 13.5 N forward
  3. What if air resistance was not negligible? How would this change the values calculated above? How would this change the…
    1. Acceleration was computed from measured values of velocity and time. Adding drag to the problem does not change these measurements, therefore acceleration does not change.
    2. Weight is determined by mass and gravity. Mass is an invariant quantity. Nothing can change it. The value of the gravitational field is determined by the location. Adding drag does not change the mass of the bird or its location, therefore weight does not change.
    3. The net force is determined by mass and acceleration. Neither of these quantities is affected by drag, therefore net force does not change.
    4. Lift balances weight. Since weight does not change, lift does not change.
    5. When there was no drag, the thrust accelerated the bird. Now that there is drag, the thrust has to accelerate the bird and overcome drag. This means that thrust is greater when drag is present than when it is not.

practice problem 2

A laboratory cart (m1 = 500 g) is pulled horizontally across a level track by a lead weight (m2 = 25 g) suspended vertically off the end of a pulley as shown in the diagram below. (Assume the string and pulley contribute negligible mass to the system and that friction is kept low enough to be ignored.)

  1. Draw a free body diagram for…
    1. the cart
    2. the weight
  2. Determine…
    1. the acceleration of the system
    2. the tension in the string


Answer it.

practice problem 3

A 100 kg wooden crate rests on a wooden ramp with an adjustable angle of inclination.
  1. Draw a free body diagram of the crate.
  2. If the angle of the ramp is set to 10°, determine…
    1. the component of the crate's weight that is perpendicular to the ramp
    2. the component of the crate's weight that is parallel to the ramp
    3. the normal force between the crate and the ramp
    4. the static friction force between the crate and the ramp
  3. At what angle will the crate just begin to slip?
  4. If the angle of the ramp is set to 30°, determine…
    1. the component of the crate's weight that is perpendicular to the ramp
    2. the component of the crate's weight that is parallel to the ramp
    3. the normal force between the crate and the ramp
    4. the kinetic friction force between the crate and the ramp
    5. the net force on the crate
    6. the acceleration of the crate



  1. Draw it

  2. This is an example of a classic physics problem that students have been solving since the Seventeenth Century.
    1. The component of the crate's weight perpendicular to the ramp is found using the cosine function. An object's weight is entirely pushing into a surface when the surface is level (a 0° angle of inclination). None of that weight is pushing into the surface when the surface is vertical, like a wall (a 90° angle of inclination). Cosine is a maximum when the angle is zero and zero when the angle is 90°. This is how the perpendicular component works.
      W = W cos θ = mg cos θ
      W = (100 kg)(9.8 m/s2)(cos 10°)
      W = 965 N
    2. The component of the crate's weight parallel to the ramp is found using the sine function. An object's weight has no sideways component on a level floor (a floor with no inclination). An object's weight is entirely parallel to a wall (a floor with a 90° inclination, in a sense). Sine is zero when the angle is zero and a maximum when the angle is 90°. This is how the parallel component works.
      W = W sin θ = mg sin θ
      W = (100 kg)(9.8 m/s2)(sin  10°)
      W = 170 N
    3. Normal forces are normal — that is, perpendicular to a tangent drawn to a curve or surface. I'll say it again, this crate isn't currently going anywhere, so all the forces perpendicular to the incline must cancel. For a static crate on an incline, the force normal to the incline equals the perpendicular component of its weight.
      N = W
      N = 965 N
    4. Friction is a sideways or lateral force — a force that is parallel to the surface of a solid. This crate isn't going anywhere, so all the forces parallel to the incline should cancel. For a static crate on an incline, the static friction force equals the parallel component of the crate's weight.
      ƒ = W
      ƒ = 965 N
  3. The component of the crate's weight parallel to the incline pulls the crate down the incline while the frictional force tries to keep it in place. Since nothing is going anywhere, these two forces must balance each other.
    ∑ F  = m a
    W − ƒ  = 0
    ƒ  = W

    As the angle of inclination increases, so to does the static friction, but it can't keep doing this forever. At some angle, the parallel component of the weight will equal the maximum static friction. Friction won't be strong enough and the crate will slip.

    ƒmax  = W
    μsmg cos θ  = mg sin θ

    Cancel the weight.

    μs cos θ = sin θ

    Do some trig.

    tan θ = μs

    Enter numbers.

    tan θ = 0.28

    Compute. The angle at which the crate just begins to slip is…

    θ = 16°

    This number is known as the critical angle (because it marks a critical value separating two types of behavior — sticking vs. sliding), angle of friction (because you gotta call it something), angle of repose (because granular materials will settle, or repose, in conical piles with this angle), or critical angle of repose (because adding grains to a pile with this angle will make it slump).

  4. This part of the problem moves us from the Seventeenth to the Eighteenth Century, because it include Newton's second law of motion in one direction. When it come to components, the choice of trig functions are still true. Only the angles have been changed to protect the innocent.
    1. The perpendicular component follows the sames rules it did in part a. Use cosine here.
      W = W cos θ = mg cos θ
      W = (100 kg)(9.8 m/s2)(cos 130°)
      W = 849 N
    2. The parallel component of the weight still uses the sine function.
      W = W sin θ = mg sin θ
      W = (100 kg)(9.8 m/s2)(sin  30°)
      W = 490 N
    3. There is no way for the crate to move perpendicular to the ramp in this scenario. The normal force must therefore equal the perpendicular component of the crate's weight.
      N = W
      N = 849 N
    4. The angle of inclination in part d is greater than the critical angle calculated in part c. Friction is no longer strong enough to keep the crate in place. The kinetic friction in this part of the problem is now really a function of the material surfaces (the coefficient of friction) and the contact forces (the normal force).
      ƒ = µkN
      ƒ = (0.17)(849 N)
      ƒ = 144 N
    5. The forces perpendicular to the surface cancel out. The forces parallel to the surface do not. One is greater than the other. The parallel component of the weight is greater than the kinetic friction force. The difference of these two is the net force, and it drags the crate down the ramp.
      F = W − ƒ
      F = 490 N − 144 N
      F = 346 N down the ramp
    6. Welcome to the Eighteenth Century. Net force and mass determine acceleration. This is Newton's second law of motion — discovered in 1666, making it the most modern part of this problem.
      a =  F  
      a =  346 N  
      100 kg  
      a = 3.46 m/s2  

practice problem 4

A pendulum can be used as an inexpensive accelerometer by a passenger in a car, airplane, roller coaster, or other vehicle. When the vehicle isn't accelerating, the pendulum will hang vertically. When the vehicle is accelerating, the pendulum will hang at an angle. Let m be the mass of the pendulum bob, be its length, a be the acceleration of the vehicle, and θ be the angle the pendulum deviates from the vertical.
  1. Draw a free body diagram for the pendulum bob.
  2. Derive an equation for acceleration of the vehicle in terms of the quantities given and known constants.


  1. The pendulum bob swings in the direction opposite the acceleration. In a sense, the bob is trying to catch up to the moving vehicle when it speeds up and is overrunning the vehicle when it slows down. Inertia in action.

    Start with a free body diagram. We have weight down and tension at an angle. Break the tension up into components in the traditional directions of horizontal and vertical.

  2. Apply Newton's second law of motion…

    ∑ F = m a

    but do it twice. (Let up and forward be the positive directions.)

    In the horizontal direction, the horizontal component of the tension is unbalanced. It is the net force.

    ∑ Fx  =  max
    T sin θ  =  ma

    In the vertical direction, we assume there is no acceleration. The upward component of the tension should balance the downward weight of the pendulum bob.

    ∑ Fy  =  may
    T cos θ − mg  =  0
    T cos θ  =  mg

    Divide these two equations.

    T sin θ  =  ma
    T cos θ mg

    Simplify using algebra and trig identities.

    tan θ =  a

    a = g tan θ

    Test the equation with a few representative values. A 0° angle indicates no acceleration, since tan 0° = 0; a 45° angle corresponds to a horizontal acceleration of 1 g, since tan 45° = 1; and a 90° angle is impossible, since tan 90° = ∞.

    Although the idea of using a pendulum to measure horizontal accelerations is a simple one, there really was no need to make one until people regularly started to move at speeds faster than a fast horse. The first pendulum accelerometer was built in 1889 by the British mechanical engineer Frederick Lanchester (1868–1946). A pencil was attached to the pendulum bob so that it could automatically draw an acceleration–time graph on a piece of paper.

    Lanchester was interested in the smoothness of braking systems on trains. In particular, he was curious about the cause of the sudden change in motion that happens right before a braking train comes to rest.

    It has been remarked that a characteristic feature of brake diagrams is the suddenness of the drop at the instant of stopping. This is a very interesting and important point, inasmuch as it is the cause of the "jerk" nearly always experienced just as a train comes to rest; it was in fact in investigating this jerk in 1888 that the idea of the pendulum accelerometer occurred to the writer. At that time it was currently supposed that the jerk was the effect of the recoil of the buffer springs after stopping; whereas a very little consideration shows that it is in reality sudden change of acceleration [emphasis original] that we recognize physiologically as "jerk," that is df/dt [for some odd reason, he chose to use f for acceleration], and not change in the direction of motion. It suggests itself in fact that the term "jerk " might well be given a scientific meaning and be defined as d3s/dt3.

    Frederick Lanchester, 1905

    The suggestion stuck.