
A force is an action upon an object that can change the object’s velocity. The SI unit of force is the newton (N). One newton is the force required to accelerate a 1 kg object by 1 m/s^{2}. Force is a vector: It has direction as well as magnitude (or strength). Weight is one example of a force. An object is in equilibrium when the net force (the vector sum of all external forces) on that object is equal to zero.

force, newton (N), pound (lb), weight, freebody diagram, normal force, center of mass, net force, equilibrium



$${F}_{1}+{F}_{2}+{F}_{3}+\cdots =0$$



Review problems and questions 


Ariel’s puppy weighs 15 lb. Heather’s puppy weighs 50 N. Whose puppy weighs more?

Ariel’s puppy weighs more than Heather’s. In order to compare the two forces, we must express them in the same units. Let’s convert 15 lb to the equivalent force in newtons:
$$15\text{lb}\times \frac{4.448\text{N}}{1\text{lb}}=66.7\text{N}$$ Since 66.7 is greater than 50, Ariel’s puppy outweighs Heather’s.


Imagine that a spacecraft has been designed, built, and launched to explore the atmosphere of the planet Mars. To enter a circular orbit around Mars, the craft has to be slowed down. The spacecraft’s reverse thruster (its “brakes”) is supposed to deliver a force of 1,000 N whenever it is fired. The thruster, however, actually delivers 1,000 lb of force. Is it too weak or too strong?

The thruster is too strong. One pound is about as strong as 4½ newtons (the precise conversion factor being 4.448). Therefore 1,000 lb is 4,448 N. That is 3,448 N stronger than required.


The acceleration due to gravity at the surface of Mars is 38% of that on Earth. If a Mars rover has a mass of 950 kg on Earth, how much does it weigh on Mars?

Answer: The rover weighs 3,500 N on Mars.
Asked:weight F_{w} of rover on Mars Given:rover mass m = 950 kg, acceleration due to gravity on Earth g = 9.8 m/s^{2}, gravity on Mars is 38% of Earth’s Relationships: F_{w} = mg Solve: $$\begin{array}{lll}{F}_{w}\hfill & =\hfill & mg\hfill \\ \hfill & =\hfill & \left(950\text{\hspace{0.17em}}\text{kg}\right)\times \left(9.8\text{\hspace{0.17em}}\text{m}/{\text{s}}^{\text{2}}\right)\times 0.38=3,500\text{\hspace{0.17em}}\text{N}\hfill \end{array}$$


A 100 kg astronaut weighs 500 N on a different planet.
 What is the astronaut’s mass on that planet?
 What is the acceleration due to gravity on that planet?

Answer:  The astronaut’s mass does not change. It is still 100 kg.
 The acceleration due to gravity on the other planet is 5 N/kg.
Asked:acceleration due to gravity g on different planet Given:astronaut mass m = 100 kg, astronaut weight F_{w} = 500 N Relationships: F_{w} = mg Solve: $${F}_{w}=mg$$Rearrange equation to solve for g:$$g=\frac{{F}_{w}}{m}=\frac{500\text{\hspace{0.17em}}\text{N}}{100\text{\hspace{0.17em}}\text{kg}}=5\text{\hspace{0.17em}}\text{N}/\text{kg}$$


A 120 lb woman sits on a 20 lb stool with her feet above the floor, as shown here. (Assume that the stool rests on a kitchen floor.)
 Is the woman in equilibrium? How about the stool?
 What is the net force on the woman? On the stool?
 Describe how to draw a freebody diagram for the woman.
 How strong is the normal force that the stool exerts upon the woman?

 The woman and the stool are both in equilibrium. Neither is moving. An object at rest is in equilibrium.
 The woman is subject to a zero net force; so is the stool. Since the woman and the stool are both in equilibrium, each must experience zero net force. The forces acting upon each of them must all cancel each other out.
 Draw an outline of the woman. Next, draw a downward arrow that starts at her center of mass. (This arrow represents her weight, which is a force.) Finally, draw an upward arrow below the woman, one that points toward her body. This arrow represents the normal force on the woman from the stool.
 The net force on the woman must be zero, since she is in equilibrium. The normal force must therefore be just as strong as her weight (120 lb, or roughly 530 N).


A student places a 15 kg box on a level floor.
 What is the net force on the box? What is the normal force on the box?
 The student now presses directly down on the box with a force of 30.0 N. What is the net force on the box? What is the normal force on the box?
 The student now pulls directly up on the box with a force of 80.0 N. What is the net force on the box? What is the normal force on the box?

 The box is at rest, so F_{net} = 0 N. The weight of the box is F_{w} = (15 kg)(9.8 m/s^{2}) = 147 N. The normal force on the box is therefore 147 N upward.
 The box is still at rest, so F_{net} = 0 N. The normal force must therefore be 30 N + 147 N = 177 N upward.
 The box is still at rest, so F_{net} = 0 N. The normal force must be 147 N − 80 N = 67 N upward.

Take a Quiz 