Concurrent and parallel forces

Summary and Practice Analyzing Simple Forces

© by William Dietsch 2005


Net forces can change the state of motion of an object.  (See section on dynamics for more discussion of net forces.)

• Forces can be thought of having lines of action, which may extend for long distances (such as in the case of gravity).

• Forces always occur in pairs.  (See the section on dynamics for a review of Newton's third law and more information about action-reaction pairs.)

• In each pair of forces, the forces in the pair act in exactly opposite directions.

• When two or more forces act simultaneously on a body at the same point of application, they are called concurrent forces.  The vector sum of the concurrent forces represents the single force which would have the same effect as the concurrent forces.  The force equal to the vector sum of the concurrent forces, but acting in an opposite direction, is called the equilibrant force.

• When all of the concurrent forces acting on a body are in balance, a state of equilibrium exists.  Bodies in equilibrium do not experience changes in their motion.

• Frequently forces act in directions which are not exactly the same as the direction of the motion of a body.  If a force acts in such a manner, it becomes necessary to resolve or split the force into components, which act at right angles to each other. This process is called resolution of forces.  The component, which then acts in exactly the direction of motion, is called the effective component and changes the body's speed.  The other component is called the ineffective component, because it does not contribute to the body's speed, but only changes the direction of motion.

Inclined planes use resolution of forces to find the component of weight, which causes a body resting on the incline to slide down the plane.  Consider a plane which makes an angle of Θ to the horizontal.  The force acting parallel to the surface of the plane and is symbolized as FP ( FP = W*sin Θ ).  The force which acts perpendicular to the plane and presses the body on to the surface is called the normal force (FN = W*cos Θ ).  The normal force is involved with friction retarding the body while the perpendicular force contributes to the acceleration of the body.

• The weight of a body is a force vector, which is assumed to have a point of application at the center of mass (center of gravity) of the body.  Symmetrical, homogeneous bodies have centers of mass at their geometric centers.  For example a uniform meter stick has a center of mass at the 50 cm mark.  Irregular bodies have their centers of mass in locations which are found experimentally by suspending the objects from three suspension points, drawing vertical lines from the suspension points and finding a point of intersection.  The center of mass of the object is the point of intersection.  The center of mass might even be outside the material object (as is the case with a tire).

Parallel forces are forces, which act on a body, with lines of action that never intersect (parallel lines of action).

• Parallel forces may cause torque (τ).  Torque is a force, which can cause twist or rotation about an axis.  Torque is the product of the force (in Newtons) multiplied times the perpendicular distance (in meters) from the point of application of the force to the pivot point.  Torque is measured in meter Newtons (m*N).  Torque causes clockwise rotation or counterclockwise rotation.

• The pivot point is that point on a body (real or virtual) about which that body rotates.  Examples of natural pivot points are elbows, knees, door hinges, and wheel hubs.

• When parallel forces act on a body, two conditions of equilibrium must be met: Translational equilibrium is when the forces are balanced (ΣF = 0), producing a state in which no linear acceleration occurs.  Rotational equilibrium occurs when the sum of the clockwise torque is equal to the sum of the counterclockwise torque (Στ = 0).

• The Law of Torques states that if a body is in a state of rotational equilibrium, the sum of the clockwise torques is equal to the sum of the counterclockwise torques.

Composition of forces

  1. Two soccer players kick a ball at the same instant. One strikes with a force of 65 N at 0° and the other 88 N at 90°.  Find the resultant force on the ball.

  2. Two children pull a wagon by exerting forces of 15 N and 18 N at the same point.  If the angle between them is 35.0°, what is the magnitude of the resultant force on the wagon?

  3. A boy and a girl carry a 12.0 kg bucket of water by holding the ends of a rope with the bucket attached at the middle.  If there is an angle of 100.0° between the two segments of the rope, what is the tension in each part?

  4. Three Men are pulling on ropes attached to a tree.  The first man exerts a force of 60.0 N at 0°, the second a force of 35 N at 90°, and the third 40 N at 150°.
    1. Find the resultant force on the tree by using the polygon method (with a scale drawing)
    2. What is the equilibrant force?

  5. A traffic light is supported by two wires, which make an angle of 140.° to each other.  If the maximum tension in each wire is 750 N, what is the maximum weight of the light they can support?

Resolution of forces

  1. A 2.00 x 103 kg car is to be held on a 20.0° incline by a rope in which the maximum tension is 8,000. N.
    1. Will the rope support the car?
    2. If the rope is released, how far will the car have moved down the incline by the time its speed reaches 35.0 m/s? (Hint: use Newton's second law to find the acceleration and a kinematics formula to find distance)

  2. A man pulls a sled by exerting a force of 35 N on a rope making an angle of 55° with the horizontal.  Find the vertical and horizontal components of the force.

  3. A Person pushes a lawn mower by exerting a 76 N force on the handle inclined at 40.0° above the horizontal.
    1. What force pushes the mower against the lawn (the vertical component)?
    2. What force moves it forward (the horizontal component)?

  4. A 20 000 N car is parked on an incline that makes an angle of 30.00° with the horizontal.  If the maximum force the brakes can withstand is 12 000 N, will the car remain at rest? (hint: if FP > 12,000 N, the car will move)

  5. Find the x and y components (i.e. the horizontal and vertical components) of an 88 N force making an angle of 22.0° with the x-axis.

  6. Two paramedics are lifting a person on a stretcher.  One of the paramedics exerts a force of 350 N at 58° above the horizontal and the other exerts a force of 410 N at 43° above the horizontal.  What is the weight of the person and the stretcher? (Diagram needed)

  7. Ms. Jones has attached a sign that has a weight of 495 N to a wall outside her office, as shown in the diagram.  Determine
    1. the magnitude of the tension in the chain and
    2. the thrust force exerted by the rod.

Parallel forces

Note: For each problem, draw and label an appropriate force diagram on a separate piece of paper.  Unless otherwise noted, the center of gravity is at the geometric center of the object.
  1. A 400.0 N child and a 300.0 N child sit on either end of a 2.00 m long seesaw.  Where along the seesaw should the pivot support be placed to ensure rotational equilibrium?

  2. Based on the information in Problem 1, suppose a 225 N child sits 0.200 m from the 400.0 N child.  Where must a 325 N child sit to maintain rotational equilibrium?

  3. A uniform meter stick, supported at the 30.0 cm mark, is balanced when a 0.50 N weight is hung at the 0.0 cm mark.  What is the weight of the meter stick?

  4. A 650 N boy and a 490 N girl sit on a 150 N porch swing that is 1.70 m long.  If the swing is supported by a chain at each end, what is the tension in each chain when the boy sits 0.750 m from the left end and the girl 0.500 m from the other?

  5. A 30.0 N fishing pole is 2.00 m long and has its center of gravity 0.350 m from the heavy end.  A fisherman holds the end of the pole in his left hand as he lifts a 100.0 N fish.  If his right hand is 0.800 m from the heavy end, how much force does he have to exert to maintain equilibrium?

  6. A uniform 2.50 N meter stick is hung from the ceiling by a single rope. A 500.0 g mass is hung at the 25.0cm mark and a 650.0 g mass at the 70.0 cm mark.
    1. What is the tension in the rope?
    2. Where is the rope attached to the meter stick?

  7. A 850 N painter stands 1.20 m from the left end of a 3.00 m scaffold supported at each end by a stepladder.  The scaffold weighs 250 N and there is a 40.0 N can of paint 0.50 m from the end opposite the painter. How much force does each stepladder exert?

  8. Two spring scales suspend a 10.0 N meter stick, one at the 8.00 cm. mark and the other at the 90.0 cm mark.  If a weight of 5.00 N is hung at the 20.0 cm mark and a weight of 17.0 N is hung at the 55.0 cm mark, what will be the reading on each scale?

  9. A pivot at the 35cm mark supports a meter stick.  It is brought into balance by a 0.85 N weight suspended at the 5.5-cm mark.  Compute the weight of the meter stick.

  10. A 3.0 m plank is suspended form both ends by ropes, has a weight of 220 N and a center of gravity 1.2 m from the left.  A 450 N man stands 2.2 m from the left and a 350 N woman stands 0.65 m from the right.  Compute the upward support on both ends.

  11. A uniform bar is 5.0 m long and weighs 56 N.  A 2.5 Kg mass is suspended on the left end and a 1.2 Kg mass is suspended 1.4 m from the right.  Compute the magnitude, direction and point of application of the single force needed to establish translational and rotational equilibrium.

  12. A park bench is 2.4  m long and is supported by legs at the ends.  The weight of the bench is 256 N and a man weighing 490 N sits 1.6 m from the left end.  Compute the upward force exerted by each leg.

  13. A uniform meter stick weighs 1.4 N, and is supported by a pivot at the 65 cm mark.  Where must a 0.85 N force be suspended to balance the stick?

  14. A 2.5-m bar weighs 87 N and has a center of gravity located 1.1 m from the left end.  A weight of 110 N is suspended 1.9 m from the left end and a weight of 55 N is suspended 1.1 m from the right.  Compute the single force needed to produce translational and rotational equilibrium.


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created and © 2005 by William Dietsch
posted & edited 4 April 2007 by D Trapp
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