Astandard four-drawer filing cabinet is 52 inches high and 15 inches wide. if it is evenly loaded, the center of gravity is at the center of the cabinet. a worker is tilting a filing cabinet to the side to clean under it.

part a

to what angle can he tilt the cabinet before it tips over?

express your answer using two significant figures.

[tex]\boxed {16}^{\circ}}[/tex]

Explanation:

Normally, the angle from one corner to the center of gravity is expressed as

[tex]Tan\theta=\frac {P}{B}[/tex] where P and B are perpendicular and base point of the cabinet.

Using the free body diagram attached then

[tex]\theta=tan ^{-1} \frac {52}{15}=73.90^{\circ}[/tex]

The angle made by the vertical line will be [tex]90-\theta[/tex] hence [tex]90^{\circ}-73.90^{\circ}=\boxed {16}^{\circ}}[/tex]

[tex]A standard four-drawer filing cabinet is 52 inches high and 15 inches wide. If it is evenly loaded,[/tex]

unlike poles attract

angle = 16.1 degree

Explanation:

given data

high = 52 inches

wide = 15 inches

solution

we know cabinet move until the CG is past vertical line of rotation point

and After CG cross line

than there weight of cabinet will be rotate

so now w eget angle by 1 corner to CG that is

angle = atan[tex]\frac{52}{15}[/tex]

angle = 73.9 degree

so now angle (x) tilt cabinet before it is tips over will be as

so as it make angle vertical that is

90 degree = x + 73.9 degree

angle (x) = 16.1 degree

x=16.10°

Explanation:

You can go through a bunch of equations, but basically the cabinet can be moved until the CG is past the vertical line of the rotation point. After the CG crosses the line the weight of the cabinet will keep the rotation going in the direction.

to Find the angle from one corner to the CG

angle =atan(52/15) = 73.90

So, to make that angle vertical...

90° = x + 73.90°

x=16.10°

double the distance = divide strength by 4