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?

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

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)

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 $566\mathrm{N}$.
.

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

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