Chapter 6.1, Problem 46E

Solution

(a)

To find: The component form of the force.

The component form of the force is equal to $〈2.41,0.65〉$ .

Given information:

In a warehouse a box is being pushed up a ${15}^{\circ }$ inclined plane with a force of $2.5$ lb, as shown in the figure.

  

Concept used:

The component form of a vector is $\overrightarrow{a}=〈\left|a\right|\cos \theta ,\left|a\right|\sin \theta 〉$ .

Calculation:

Apply the above concept and find the component form of the force vector $\overrightarrow{v}$ below:

  $\begin{array}{c}\overrightarrow{v}=〈2.5\cos {15}^{\circ },2.5\sin {15}^{\circ }〉\\ =〈2.41,0.65〉\end{array}$

(b)

To find: The horizontal component and vertical component of the force.

The horizontal component is equal to $2.41$ and the vertical component is equal to $0.65$ .

Given information:

In a warehouse a box is being pushed up a ${15}^{\circ }$ inclined plane with a force of $2.5$ lb, as shown in the figure.

  

Concept used:

The horizontal component of a vector is $\left|a\right|\cos \theta$ .

The vertical component of a vector is $\left|a\right|\sin \theta$ .

Calculation:

Apply the above concept and find the horizontal component and vertical component below:

The horizontal component of the force is $2.5\cos {15}^{\circ }=2.41$ .

The vertical component of the force is $2.5\sin {15}^{\circ }=0.65$ .