Question

Two horizontal forces,

${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{}\mathbf{and}\mathbf{}{\overrightarrow{\mathbf{F}}}_{\mathbf{2}}\mathbf{=}\mathbf{‐}\mathbf{(}\mathbf{1}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{‐}\mathbf{(}\mathbf{2}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}$

pull a banana split across a frictionlesslunch counter. Without using acalculator, determine which of thevectors in the free-body diagram ofFig. 5-20 best represent (a) ${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}$and(b) ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}$ . What is the net-force componentalong (c) the xaxis and (d) the yaxis? Into which quadrants do (e) thenet-force vector and (f) the split’s accelerationvector point?

Short answer

a) Line 5 represents Vector ${\mathrm{F}}_1$

b) Line 7 represents Vector ${\mathrm{F}}_2$

c) X component of the net force is 2 N

d) Y component of the net force is -6 N

e) Fourth

f) Fourth

Step-by-step solution

Given information

Figure 5-20 is given.

To understand the concept

The problem is based on Newton’s second law of motion. The law states that the acceleration of an object is dependent on the net force acting upon the object and the mass of the object. Using Newton’s second law of motion and vector resolution, the answers to the above questions can be found.

Formula:${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{Ma}$

(a) To find Force Vector F1

From graph line 5 represents vector $F_1$as it has a positive x component and negative y component. Also, the x components and y components are longer than the components in vector 6.

(b) To find Force Vector F2

From graph line 7 represents vector ${\mathrm{F}}_2$as it has both x and y components negative. Also, both components are substantially smaller than vector 8.

(c) To find X component of the net force

X component of the net force is the addition of x components of vector ${\mathrm{F}}_1$and${\mathrm{F}}_2$

So x component is,

$$\begin{gathered} \mathrm{F}=3\mathrm{N}-1\mathrm{N} \\ =2\mathrm{N} \end{gathered}$$

(d) To find the Y component of the net force

y component of the net force is the addition of x components of vector ${\mathrm{F}}_1$and${\mathrm{F}}_2$

So y component is

$$\begin{gathered} \mathrm{F}=-2\mathrm{N}-4\mathrm{N} \\ =-6\mathrm{N} \end{gathered}$$

(e) To find the net force vector

Since the X component is positive and the Y component is negative, the resultant force is in the fourth quadrant.

(f) To find the acceleration vector

Net force and acceleration would be in the same direction;therefore, the acceleration would also be in the fourth quadrant.