Normal Force - AP Physics C: Electricity and Magnetism

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Question

Weightlessness is experienced when the normal force equals .

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Answer

"Weightlessness" is analogous to free-fall (neglecting air resistance), during which the only force on an object is the force of gravity. When normal force becomes zero, the object loses physical contact with the surface.

When normal force is equal to the force of gravity, the object is in equilibrium and the net vertical force is zero. This results in zero acceleration, but does not result in "weightlessness." Mass is not a force, and negative force values merely indicate relative direction (assuming correct calculations)

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Question

A book of mass 1kg is held to a vertical wall by a person's hand applying a 20N force directly toward the wall. The wall has a static friction coefficient of 0.3 and a kinetic friction coefficient of 0.2. With the book held at rest, what the is frictional force keeping the book from sliding down the wall?

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Answer

Frictional force is calculated using the normal force and the friction coefficient. Since the book is at rest (not in motion), we use the coefficient of static friction.

$F_f=\mu_sF_N$

The normal force will be equal and opposite the force pushing directly into the surface; in this case, that means the normal force will be 20N.

Solve for the force of friction:

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Question

$\begin{align}&\text{A block of mass }190kg\text{ rests on an incline of }33^{\circ}\&\text{It is also partially supported by a rope with a tension of }559.17N\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

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Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), the force of the rope supporting it, a normal force,}\&\text{and since it’s on an incline, a frictional force to balance}\&\text{it all out. Notice that the tension in the rope mitigates some}\&\text{of the gravitational force (as illustrated in the upper left}\&\text{corner) to lessen what is applied to the slope:}\&F_{mass}=mg\&F_{normal}=(F_{mass}-F_{rope})cos(a)\&F_{friction}=(F_{mass}-F_{rope})sin(a)\&F_{normal}=(190kg(9.81\frac{m}{s^2})-559.17N)(cos(33^{\circ}))\&F_{normal}=1094.24N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }233kg\text{ rests on an incline of }81^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

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Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=233kg(9.81\frac{m}{s^2})(cos(81^{\circ}))\&F_{normal}=357.57N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }187kg\text{ rests on an incline of }83^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

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Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=187kg(9.81\frac{m}{s^2})(cos(83^{\circ}))\&F_{normal}=223.57N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }267kg\text{ rests on an incline of }49^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

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Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=267kg(9.81\frac{m}{s^2})(cos(49^{\circ}))\&F_{normal}=1718.4N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }112kg\text{ rests on an incline of }18^{\circ}\&\text{It is also partially supported by a rope.}\&\text{The normal force acting on the block is }417.98N\&\text{What is the tension felt in the rope?}\&g=9.81\frac{m}{s^2}\end{align}$

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Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), the force of the rope supporting it, a normal force,}\&\text{and since it’s on an incline, a frictional force to balance}\&\text{it all out. Notice that the tension in the rope mitigates some}\&\text{of the gravitational force (as illustrated in the upper left}\&\text{corner) to lessen what is applied to the slope:}\&F_{mass}=mg\&F_{normal}=(F_{mass}-F_{rope})cos(a)\&F_{rope}=F_{mass}-\frac{F_{normal}}{cos(a)}\&F_{rope}=1098.72N-\frac{417.98N}{cos(18^{\circ})}\&F_{rope}=659.23N\end{align}$

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Question

Weightlessness is experienced when the normal force equals .

Tap to reveal answer

Answer

"Weightlessness" is analogous to free-fall (neglecting air resistance), during which the only force on an object is the force of gravity. When normal force becomes zero, the object loses physical contact with the surface.

When normal force is equal to the force of gravity, the object is in equilibrium and the net vertical force is zero. This results in zero acceleration, but does not result in "weightlessness." Mass is not a force, and negative force values merely indicate relative direction (assuming correct calculations)

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Question

A book of mass 1kg is held to a vertical wall by a person's hand applying a 20N force directly toward the wall. The wall has a static friction coefficient of 0.3 and a kinetic friction coefficient of 0.2. With the book held at rest, what the is frictional force keeping the book from sliding down the wall?

Tap to reveal answer

Answer

Frictional force is calculated using the normal force and the friction coefficient. Since the book is at rest (not in motion), we use the coefficient of static friction.

$F_f=\mu_sF_N$

The normal force will be equal and opposite the force pushing directly into the surface; in this case, that means the normal force will be 20N.

Solve for the force of friction:

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Question

$\begin{align}&\text{A block of mass }190kg\text{ rests on an incline of }33^{\circ}\&\text{It is also partially supported by a rope with a tension of }559.17N\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), the force of the rope supporting it, a normal force,}\&\text{and since it’s on an incline, a frictional force to balance}\&\text{it all out. Notice that the tension in the rope mitigates some}\&\text{of the gravitational force (as illustrated in the upper left}\&\text{corner) to lessen what is applied to the slope:}\&F_{mass}=mg\&F_{normal}=(F_{mass}-F_{rope})cos(a)\&F_{friction}=(F_{mass}-F_{rope})sin(a)\&F_{normal}=(190kg(9.81\frac{m}{s^2})-559.17N)(cos(33^{\circ}))\&F_{normal}=1094.24N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }233kg\text{ rests on an incline of }81^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=233kg(9.81\frac{m}{s^2})(cos(81^{\circ}))\&F_{normal}=357.57N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }187kg\text{ rests on an incline of }83^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=187kg(9.81\frac{m}{s^2})(cos(83^{\circ}))\&F_{normal}=223.57N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }267kg\text{ rests on an incline of }49^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

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Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=267kg(9.81\frac{m}{s^2})(cos(49^{\circ}))\&F_{normal}=1718.4N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }112kg\text{ rests on an incline of }18^{\circ}\&\text{It is also partially supported by a rope.}\&\text{The normal force acting on the block is }417.98N\&\text{What is the tension felt in the rope?}\&g=9.81\frac{m}{s^2}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), the force of the rope supporting it, a normal force,}\&\text{and since it’s on an incline, a frictional force to balance}\&\text{it all out. Notice that the tension in the rope mitigates some}\&\text{of the gravitational force (as illustrated in the upper left}\&\text{corner) to lessen what is applied to the slope:}\&F_{mass}=mg\&F_{normal}=(F_{mass}-F_{rope})cos(a)\&F_{rope}=F_{mass}-\frac{F_{normal}}{cos(a)}\&F_{rope}=1098.72N-\frac{417.98N}{cos(18^{\circ})}\&F_{rope}=659.23N\end{align}$

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Question

A book of mass 1kg is held to a vertical wall by a person's hand applying a 20N force directly toward the wall. The wall has a static friction coefficient of 0.3 and a kinetic friction coefficient of 0.2. With the book held at rest, what the is frictional force keeping the book from sliding down the wall?

Tap to reveal answer

Answer

Frictional force is calculated using the normal force and the friction coefficient. Since the book is at rest (not in motion), we use the coefficient of static friction.

$F_f=\mu_sF_N$

The normal force will be equal and opposite the force pushing directly into the surface; in this case, that means the normal force will be 20N.

Solve for the force of friction:

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Question

Weightlessness is experienced when the normal force equals .

Tap to reveal answer

Answer

"Weightlessness" is analogous to free-fall (neglecting air resistance), during which the only force on an object is the force of gravity. When normal force becomes zero, the object loses physical contact with the surface.

When normal force is equal to the force of gravity, the object is in equilibrium and the net vertical force is zero. This results in zero acceleration, but does not result in "weightlessness." Mass is not a force, and negative force values merely indicate relative direction (assuming correct calculations)

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Question

$\begin{align}&\text{A block of mass }190kg\text{ rests on an incline of }33^{\circ}\&\text{It is also partially supported by a rope with a tension of }559.17N\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), the force of the rope supporting it, a normal force,}\&\text{and since it’s on an incline, a frictional force to balance}\&\text{it all out. Notice that the tension in the rope mitigates some}\&\text{of the gravitational force (as illustrated in the upper left}\&\text{corner) to lessen what is applied to the slope:}\&F_{mass}=mg\&F_{normal}=(F_{mass}-F_{rope})cos(a)\&F_{friction}=(F_{mass}-F_{rope})sin(a)\&F_{normal}=(190kg(9.81\frac{m}{s^2})-559.17N)(cos(33^{\circ}))\&F_{normal}=1094.24N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }233kg\text{ rests on an incline of }81^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=233kg(9.81\frac{m}{s^2})(cos(81^{\circ}))\&F_{normal}=357.57N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }187kg\text{ rests on an incline of }83^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=187kg(9.81\frac{m}{s^2})(cos(83^{\circ}))\&F_{normal}=223.57N\end{align*}$

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Question

$\begin{align}&\text{A block of mass }267kg\text{ rests on an incline of }49^{\circ}\&\text{What is the normal force acting on the block from the surface?}\&g=9.81\frac{m}{s^2}\end{align}$

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Answer

$\begin{align}&\text{To solve a problem of force balance, a good idea is to draw}\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\&\text{of forces acting on it must be equal to zero. The forces acting}\&\text{on it are gravity (in the form of the mass times the gravitational}\&\text{constant), a normal force, and since it’s on an incline, a frictional}\&\text{force to balance it all out:}\&F_{mass}=mg\&F_{normal}=mgcos(a)\&F_{friction}=mgsin(a)\&F_{normal}=267kg(9.81\frac{m}{s^2})(cos(49^{\circ}))\&F_{normal}=1718.4N\end{align*}$

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