Question

In Figure 5-40, a crate of mass $\mathit{m}\mathbf{=}\mathbf{100}\mathbf{}\mathit{k}\mathit{g}$is pushed at constant speed up a frictionless ramp$\mathit{(}\mathit{\theta }\mathit{=}\mathit{30}\mathit{.}\mathit{0}\mathit{°}\mathit{)}$by a horizontal force$\overrightarrow{F}$. (a) What are the magnitudes ofand (b) What are the magnitudes of the force on the crate from the ramp?

Short answer

1. The magnitude of $\overrightarrow{\mathrm{F}}=566\mathrm{N}$
2. The force on the crate from the ramp as normal force$\mathrm{N}=1.13\times {10}^3\mathrm{N}$

Step-by-step solution

Given information

• The mass of the crate is$\mathrm{m}=100\mathrm{kg}$
• The angle of inclination by the crate is $\theta =30°$

Understanding the concept of force and free body diagram

Newton’s second law states that the product of mass and acceleration is equal to the force acting on the body. All the forces acting on the body can be shown by the free body diagram.

The free-body diagram of a crate on a ramp has to be drawn. By using Newton’s second law, the magnitude of Fand the force on the crate from the ramp which is normal force Ncan be calculated. The crate is moving upward at a constant speed hence the acceleration of the crate is zero. Also, use the concept of components resolution.

Formula:

${\overrightarrow{\mathbf{F}}}_{\mathbf{net}}\mathbf{}\mathbf{=}\mathbf{m}\overrightarrow{\mathbf{a}}$ (1)

Draw the free body diagram

a) Calculate the magnitude of  F→

The crate is moving with constant velocity; hence it has zero acceleration. We can use sign convention according to the motion of the crate as shown in the free body diagram.

Use Newton’s second law along the x-axis as follows:

$$\begin{gathered} \mathrm{F}\mathrm{cos\theta }-\mathrm{mg}\mathrm{sin\theta }=0 \\ \mathrm{F}\mathrm{cos\theta }-\mathrm{mg}\mathrm{sin\theta } \\ \mathrm{F}=\frac{\mathrm{mg}\mathrm{sin\theta }}{\mathrm{cos\theta }} \\ \mathrm{F}=\mathrm{mgtan\theta } \end{gathered}$$

Substitute the values of mass, gravitational acceleration, and angle in the above equation to calculate the force.

$$\begin{gathered} F=100\mathrm{kg}\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\tan 30° \\ =566\mathrm{N} \end{gathered}$$

Hence, the magnitude of the force is $566\mathrm{N}$.
.

b) Calculate the force on the crate from the ramp

Use Newton’s second law along the y axis as,

$$\begin{gathered} \mathrm{N}-\mathrm{Fsin\theta }-\mathrm{mg}\mathrm{cos\theta }=0 \\ \mathrm{N}=\mathrm{Fsin\theta }+\mathrm{mg}\mathrm{cos\theta } \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{normal}\mathrm{reaction}. \\ \mathrm{N}=\left(566\mathrm{N}\times \sin 30°\right)+\left(100\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \cos 30°\right) \\ =1.13\times {10}^3\mathrm{N} \\ \mathrm{Hence},\mathrm{the}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{normal}\mathrm{force}\mathrm{is}1.13\times {10}^3\mathrm{N} \end{gathered}$$