9781118230725.pdf

110 CHAPTER 5 FORCE AND MOTION—I

Sample Problem 5.04 Cord accelerates box up a ramp

Many students consider problems involving ramps (inclined

planes) to be especially hard. The difficulty is probably visual

because we work with (a) a tilted coordinate system and (b) the

components of the gravitational force, not the full force. Here is

a typical example with all the tilting and angles explained. (In

WileyPLUS, the figure is available as an animation with

voiceover.) In spite of the tilt, the key idea is to apply Newton’s

second law to the axis along which the motion occurs.

In Fig. 5-15a, a cord pulls a box of sea biscuits up along a

frictionless plane inclined at angle u30.0. The box has

massm5.00 kg, and the force from the cord has magni-

tudeT25.0 N. What is the box’s acceleration aalong the

inclined plane?

KEY IDEA

The acceleration along the plane is set by the force compo-

nents along the plane (not by force components perpendi-

cular to the plane), as expressed by Newton’s second law

(Eq. 5-1).

Calculations:We need to write Newton’s second law for

motion along an axis. Because the box moves along the in-

clined plane, placing an xaxis along the plane seems reason-

able (Fig. 5-15b). (There is nothing wrong with using our

usual coordinate system, but the expressions for compo-

nents would be a lot messier because of the misalignment of

thexaxis with the motion.)

After choosing a coordinate system, we draw a free-

body diagram with a dot representing the box (Fig. 5-15b).

Then we draw all the vectors for the forces acting on the box,

with the tails of the vectors anchored on the dot. (Drawing

the vectors willy-nilly on the diagram can easily lead to errors,

especially on exams, so always anchor the tails.)

Force Tfrom the cord is up the plane and has magni-

:

θ

y

FN x

Fg

T

(b)

Cord

θ

(a)

The box accelerates.

Normal force

Cord's pull

Gravitational

force

x

T

mgsinθ

(g)(h)

θ mgcos

mg

θ

mgsinθ

y

FN x

(i)

mgcosθ

The net of these

forces determines

the acceleration.

These forces

merely balance.

(e)

Fg

(c) (d)

90°−

θ

θ 90°−θ

θ θ

(f)

θ

This is a right

triangle.

Parallel

component of

Fg

This is also.

Hypotenuse

Adjacent leg

(use cos )θ

Opposite leg

(use sin )θ

Perpendicular

component of

Fg

Figure 5-15(a) A box is pulled up a plane by a

cord. (b) The three forces acting on the

box: the cord’s force the gravitational force

and the normal force (c)–(i) Finding

the force components along the plane and

perpendicular to it.InWileyPLUS, this figure

is available as an animation with voiceover.

FN

:

Fg.

:

,

T

:

,

A

tudeT25.0 N. The gravitational force Fgis downward (of

: