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# Using Force Diagrams

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## Using Force Diagrams Answer Key

Instructions: Read each question carefully. Choose the answer that best fits the question. Short answer response questions must be responded to in complete sentences. If the question involves calculations, you must show all your math work.

1.
Explain the normal force. Give an example as part of your explanation.
• The normal force is the contact force, exerted perpendicular to the contact surface, on an object by, for example, the surface of a floor or wall, preventing the object from penetrating that surface.
2.
Given a force diagram like the one shown, if the object is not moving, what do you know about the forces at A and B?
1. $F_"norm"$ is less than $F_g$ because the object stays down on the surface.
2. $F_g$ is always greater than $F_"norm"$ when an object is not moving.
3. $F_"norm"=F_g$ when an object is not moving.
4. $F_"norm"=-(F_g)$ making the $F_"net"=0$ when an object is not moving.
3.
If a net force diagram is drawn, such as shown, wherein the $F_"norm"" at A "=-(F_g)" at C"$, what do you know about the object?
1. The object is accelerating in the direction of the $F_"app"" at B"$ because there is no opposing force to B, and $F_"app"" at "B>0N$.
2. The object is accelerating in the opposite direction of $F_"app"" at B"$ because forces act in equal but opposite pairs.
3. The object is not moving because the $F_"net"=F_"norm"+F_g=0N$.
4. The object is accelerating in the direction of the $F_"app"" at B"$ because the opposing $F_"fric"$ must be negligible, and $F_"app"" at "B>0N$.
4.
Calculate the net force acting on this object, and determine, if it is moving, in which direction. The normal force is 500N, the force at C is -372N, the resistive force of friction is -32N, and the force at B is 32N.
1. $F_"net"=-95N$; moving to the left
2. $F_"net"=872N$; at rest
3. $F_"net"=+95N$; moving upwards
4. impossible to determine without knowing the object's mass
5.
If a $14kg$ object is resting on the floor, calculate the normal force acting on the object.
• Given: $m=14kg$; $g=-9.8m/s^2$; and $F_g=m*g$

$F_g=(14kg)(-9.8m/s^2)=-137.2N$
When an object is at rest, $F_"norm"=-(F_g)$.

$F_"norm"=-(F_g)=+137.2N$
6.
Draw a force diagram of a 37kg object at rest on a table. Show your calculations in determining force values for A and B.

• Given: $m=37kg$ and $g=-9.8m/(s^2)$

$B=F_g=m*g=(37kg)(-9.8m/(s^2))=-362.6N$

For an object at rest, $F_"norm"+F_g=F_"net"=0$, so $F_"norm"=+362.6N$.
7.
If a net-force diagram is drawn, such as shown, wherein the $F_"norm"" at A "=-(F_g)" at B"$, what do you know about the object if the magnitude of $F_"fric"" at "C>0$?
• The object is decelerating to the right because the opposing $F_"fric"" at C"$ must be greater than the $F_"app"$.
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