Question #6178

A box of books is initially at rest a distance D = 0.600 m from the end of a wooden board. The coefficient of static friction between the box and the board is μs = 0.324, and the coefficient of kinetic friction is μk = 0.248. The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board.

Expert's answer

Let m be the mass of the box, and suppose that the angle of the board is

alpha.

Notice that there are two forces acting on the box:

1)

gravitation force F=mg

2) the board reaction force N = mg cos(alpha),

3) the force of static friction Fs

see figure in the attached

file.

Then the sum F + N is directed along the board, and is equal to

T=mg sin(alpha)

The friction force is opposite to T whenever

T < μs N

Substituing the values of T and N we get:

mg

sin(alpha) < μs mg cos(alpha)

whence

tan(alpha) < μs

Thus

the will box start to move when

tan(alpha) = μs

Whence

cos(alpha) = 1/square_root(1+μs^2)

sin(alpha) =

μs/square_root(1+μs^2)

Suppose now that the box is

moved.

Then instead of the force of static friction Fs, there is a force of

kinetic friction equal to

Fk = μk N = μk mg cos(alpha)

So the

resulting force acting on the box is

F = mg sin (alpha) - μk mg

cos(alpha),

where

tan(alpha) = μs

cos(alpha) =

1/square_root(1+μs^2)

sin(alpha) = μs/square_root(1+μs^2)

Whence

F = mg μs/square_root(1+μs^2) - mg μk/square_root(1+μs^2)=

=

mg(μs-μk)/square_root(1+μs^2)

Thus the box moves with acceleration

a = F/m = g(μs-μk)/square_root(1+μs^2)

with initial zero velocity.

We need

to find its velocity at the distance d = 0.6 m

We have that

v =

at,

and

d = at^2 /2, where t is the corresponding time,

whence

t = square_root(2d/a)

and so

v = at = a square_root(2d/a)

=

square_root(2da) =

= square_root( 2dg(μs-μk)/square_root(1+μs^2)

)

Substituting the values:

d=0.6

g=9.8

μs = 0.324

μk =

0.248

we get

v = square_root( 2* 0.6 * 9.8 (0.324 -

0.248)/square_root(1+0.324^2) )=

= 0.92209 m/s

alpha.

Notice that there are two forces acting on the box:

1)

gravitation force F=mg

2) the board reaction force N = mg cos(alpha),

3) the force of static friction Fs

see figure in the attached

file.

Then the sum F + N is directed along the board, and is equal to

T=mg sin(alpha)

The friction force is opposite to T whenever

T < μs N

Substituing the values of T and N we get:

mg

sin(alpha) < μs mg cos(alpha)

whence

tan(alpha) < μs

Thus

the will box start to move when

tan(alpha) = μs

Whence

cos(alpha) = 1/square_root(1+μs^2)

sin(alpha) =

μs/square_root(1+μs^2)

Suppose now that the box is

moved.

Then instead of the force of static friction Fs, there is a force of

kinetic friction equal to

Fk = μk N = μk mg cos(alpha)

So the

resulting force acting on the box is

F = mg sin (alpha) - μk mg

cos(alpha),

where

tan(alpha) = μs

cos(alpha) =

1/square_root(1+μs^2)

sin(alpha) = μs/square_root(1+μs^2)

Whence

F = mg μs/square_root(1+μs^2) - mg μk/square_root(1+μs^2)=

=

mg(μs-μk)/square_root(1+μs^2)

Thus the box moves with acceleration

a = F/m = g(μs-μk)/square_root(1+μs^2)

with initial zero velocity.

We need

to find its velocity at the distance d = 0.6 m

We have that

v =

at,

and

d = at^2 /2, where t is the corresponding time,

whence

t = square_root(2d/a)

and so

v = at = a square_root(2d/a)

=

square_root(2da) =

= square_root( 2dg(μs-μk)/square_root(1+μs^2)

)

Substituting the values:

d=0.6

g=9.8

μs = 0.324

μk =

0.248

we get

v = square_root( 2* 0.6 * 9.8 (0.324 -

0.248)/square_root(1+0.324^2) )=

= 0.92209 m/s

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