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Force: Problem Set with Solutions

This problem set is designed to accompany the PhysicsLibrary entry force. The exercises are stated first for self-study. Complete, worked solutions follow afterward.

Exercises

Problem 1. Net force from perpendicular forces. Figure 1 shows a ring pulled by two perpendicular forces, F1 = 6 N to the right and F2 = 8 N upward. Determine the magnitude and direction of the net force.

PIC

Figure 1: Two perpendicular forces acting on a ring.

Problem 2. Horizontal motion with kinetic friction. A 4.0 kg block is pushed across a horizontal floor by a constant horizontal force of 25 N, as shown in Figure 2. The coefficient of kinetic friction is μk = 0.30. Find (a) the Normal force, (b) the kinetic friction force, and (c) the horizontal acceleration of the block.

PIC

Figure 2: Free-body diagram for a block pushed on a rough horizontal surface.

Problem 3. Static friction on an incline. A 5.0 kg block rests on a 25∘ incline, as in Figure 3. Assume the block is just on the verge of slipping downward. Find the minimum coefficient of static friction required to keep it at rest.

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Figure 3: Block on an incline with Weight, normal force, and static friction.

Problem 4. An ideal Atwood Machine. In the system shown in Figure 4, m1 = 2.0 kg and m2 = 3.0 kg. The string and Pulley are ideal. Find (a) the magnitude of the acceleration and (b) the string Tension.

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Figure 4: Free-body diagram for an ideal Atwood machine.

Problem 5. Spring force and simple horizontal motion. A 0.40 kg block is attached to a horizontal spring with spring constant k = 80 N∕m. The spring is stretched by x = 0.15 m and released from rest, as in Figure 5. Find (a) the magnitude of the initial spring force and (b) the initial acceleration of the block.

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Figure 5: Block attached to a stretched horizontal spring.

Problem 6. Centripetal force on a flat curve. A 1200 kg CAR travels at 12 m∕s around a flat circular curve of radius 50 m, as sketched in Figure 6. Find (a) the required centripetal force and (b) the minimum coefficient of static friction required if static friction provides the centripetal force.

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Figure 6: Car undergoing uniform circular motion on a flat curve.

Worked solutions

Solution 1. Because the two forces are perpendicular, the magnitude of the net force is found from the Pythagorean theorem:

       ∘ ---------  √ -------
Fnet =   F 21 + F22 =   62 + 82 = 10 N.

The direction above the positive horizontal axis is

          (  )
𝜃 = tan− 1  8- ≈  53.1∘.
            6

Therefore the net force is

|------------------------------------------------|
|Fnet = 10 N,     𝜃 ≈ 53.1∘ above the horizontal. |
-------------------------------------------------

Solution 2. On a horizontal surface with no vertical acceleration,

N =  mg =  (4.0)(9.8) = 39.2N.

The kinetic friction force is

f =  μ N  = (0.30)(39.2) = 11.76 N ≈  11.8N.
 k    k

The net horizontal force is

Fnet,x = Fpush − fk = 25.0 − 11.76 = 13.24 N.

Thus the acceleration is

    Fnet,x   13.24
a = ------=  ------= 3.31 m ∕s2.
      m       4.0

So the answers are

-------------------------------------------------
|                                              2 |
-N--=-39.2N,------fk ≈-11.8N,-----a-≈-3.31-m-∕s-.|

Solution 3. Resolve the weight into components parallel and perpendicular to the incline:

mg  sin 𝜃  down  the incline,     mg cos𝜃   into the incline.

At the threshold of slipping, static friction takes its maximum value,

fs,max = μsN.

Since the block is in equilibrium,

mg  sin 𝜃 = fs,max =  μsN,     N  = mg  cos𝜃.

Therefore

     mg--sin-𝜃                 ∘
μs = mg  cos𝜃 = tan 𝜃 = tan 25 ≈  0.466.

Hence the minimum coefficient is

|------------|
|μmin≈  0.47.|
--s-----------

Solution 4. Because m2 > m1, mass m2 moves downward and m1 moves upward. Applying Newton’s second law to each mass gives

T − m1g  = m1a,

m2g − T  = m2a.

Adding the equations,

(m2  − m1 )g = (m1 + m2 )a,

so

    (m2--−-m1-)g   (3.0-−-2.0)(9.8)            2
a =   m  +  m    =     3.0 + 2.0    = 1.96 m ∕s .
        1    2

Now solve for the tension:

T  = m  (g + a) = 2.0(9.8 + 1.96) = 23.52N.
       1

Therefore

|------------------------------|
|            2                 |
-a =-1.96-m-∕s-,----T--≈-23.5N.--

Solution 5. Hooke’s law gives the magnitude of the spring force:

Fs =  kx = (80)(0.15) = 12 N.

Initially this is the net horizontal force on the block, so

    Fs     12           2
a = ---=  ---- = 30 m ∕s .
    m     0.40

The spring force and acceleration are directed toward equilibrium. Thus

|----------------------------|
|F  = 12 N,     a = 30 m ∕s2.|
--s--------------------------

Solution 6. The required centripetal force is

         2             2
Fc =  mv---= (1200-)(12--)=  3456 N.
       r          50

On a flat curve, the centripetal force is supplied by static friction. The normal force is

N  = mg  = (1200)(9.8) = 11760 N.

Thus the minimum coefficient of static friction is

       Fc    3456
μmsin = ---=  ------≈  0.294.
       N     11760

Therefore

|------------------------------|
|                   min         |
-Fc-=-3456-N,-----μ-s--≈--0.29.-

Study notes

These six problems illustrate several core ideas related to force:

  • forces add vectorially;
  • free-body diagrams help isolate all forces acting on a body;
  • Friction forces must be modeled carefully, distinguishing static and kinetic friction;
  • Newton’s second law connects force to acceleration;
  • spring forces are modeled by Hooke’s law;
  • centripetal force is not a new interaction but the inward net force needed for circular motion.

References

[1]   Daniel Kleppner and Robert J. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge University Press, 2014.

[2]   John R. Taylor, Classical Mechanics, University Science Books, 2005.

[3]   David Halliday, Robert Resnick, and Jearl Walker, Fundamentals of Physics, 10th ed., Wiley, 2013.


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Cross-references: static, Friction, Hooke's law, mass, equilibrium, theorem, uniform circular motion, CAR, centripetal force, Tension, Pulley, system, Atwood Machine, Weight, static friction, Free-body diagram, acceleration, Normal, kinetic friction, magnitude, force

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