Chapter 4, Problem 7WUE

Calculate the normal force on a 15.0 kg block in the following circumstances: (a) The block is resting on a level surface. (b) The block is resting on a surface tilted up at a 30.0° angle with respect to the horizontal. (c) The block is resting on the floor of an elevator that is accelerating upwards at 3.00 m./s². (d) The block is on a level surface and a force of 125 N is exerted on it at an angle of 30.0° above the horizontal. (Sec Section 1.5.)

To determine

The normal force when a block is resting on a horizontal level surface.

Answer to Problem 7WUE

Solution:

The normal force acting on the block is $147\text{N}$

Explanation of Solution

Given Info:

The mass of the block is $\text{15}\text{.0}\text{kg}$ and acceleration due to gravity is $9.80\text{m}/{\text{s}}^{\text{2}}$.

Write the formula to calculate normal force.

$n=mg$

• n is the normal force
• m is the mass of the block
• g is the acceleration due to gravity

Substitute $\text{15}\text{.0}\text{kg}$ for m and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate n.

$\begin{array}{c}n=\left(15.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =147\text{N}\end{array}$

Conclusion:

Therefore, the normal force acting on the block is $147\text{N}$

To determine

The normal force if the block is resting on a surface tilted up at a $30.0°$ angle with respect to the horizontal.

Answer to Problem 7WUE

Solution:

The normal force on the block is $127\text{N}$.

Explanation of Solution

Given Info:

When the surface is tilted the normal force would be one of the components of the weight of the block.

Write the formula to calculate the normal force on the block.

$n=mg\cos 30.0°$

Substitute $\text{15}\text{.0}\text{kg}$ for m and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate n.

$\begin{array}{c}n=\left(15.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\cos 30.0°\\ =\left(147\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(0.866\right)\\ \approx 127\text{N}\end{array}$

Conclusion:

Therefore, the normal force on the block is $127\text{N}$.

To determine

The normal force on a block which is resting on the floor of an elevator that accelerating upwards at $3.00\text{m}/{\text{s}}^{\text{2}}$.

Answer to Problem 7WUE

Solution:

The normal force on the block on the floor of elevator is $192\text{N}$

Explanation of Solution

Given Info:

The normal force on the block on the floor of an elevator is $3.00\text{m}/{\text{s}}^{\text{2}}$.

Write the equation of motion of the block according to Newton’s second law.

$n-mg=ma$

• a is the acceleration of the elevator

Write the formula to calculate normal force from the above expression.

$n=m\left(g+a\right)$

Substitute $\text{15}\text{.0}\text{kg}$ for m, $3.00\text{m}/{\text{s}}^{\text{2}}$ for a and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate n.

$\begin{array}{c}n=\left(\text{15}\text{.0}\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}+3.00\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(\text{15}\text{.0}\text{kg}\right)\left(12.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =192\text{N}\end{array}$

Conclusion:

Therefore, the normal force on the block on the floor of elevator is $192\text{N}$.

To determine

The normal force on a block which is encounter by an external force acting at an angle of $30.0°$ above the horizontal.

Answer to Problem 7WUE

Solution:

The normal force on the block is $\text{84}\text{.5}\text{N}$.

Explanation of Solution

Given Info:

In this case due to the presence of an external force, apart from the weight there would be one more component for downward forces.

Write the equation of motion of the block according to Newton’s second law of motion.

$n-mg+F\sin \theta =0$

• F is the magnitude of external force acting on the block.
• $\theta$ is the angle of inclination

Rewrite the above equation to calculate normal force.

$n=mg-F\sin \theta$

Substitute $\text{15}\text{.0}\text{kg}$ for m, $125\text{N}$ for F, $30.0°$ for $\theta$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g to calculate n.

$\begin{array}{c}n=\left(\text{15}\text{.0}\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)-\left(125\text{N}\right)\left(\sin 30.0°\right)\\ =\left(147.0\text{N}\right)-(62.5\text{N)}\\ \text{=84}\text{.5}\text{N}\end{array}$

Conclusion:

Therefore, the normal force on the block under the external force is $\text{84}\text{.5}\text{N}$