MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

8.01 Physics 8.01

W02D2_1 Table Problem: Free Body Force Diagrams Solution

A block of mass m is sliding down an inclined plane with angle ! . There is kinetic friction between the block and the inclined plane with coefficient µk . The gravitational force on the block is directed in the downward vertical direction.

a) Draw a free-body (force) diagram for the block. Clearly identify your forces with whatever symbols you find appropriate.

b) For the coordinate system with unit vectors i pointing horizontally to the right

and j pointing vertically upward, what are the vector components of the sum of the forces in the i -direction and the j -direction?

c) For the coordinate system with unit vectors b pointing along the inclined plane

and c pointing normal (perpendicular) to the plane as shown in the figure, what are the vector components of the sum of the forces in the b -direction and the c -direction?

Solution: There are three forces acting on the block, kinetic friction

!fk opposing the

sliding motion, gravity m!g pointing downward, and the normal force

!N perpendicular

to the contact surface. The free body force diagram on the block is shown in the figure below.

a) Let’s first choose unit vectors i pointing horizontally to the right and j pointing vertically upward as shown in the figure below.

With these unit vectors, the vector decomposition of the forces are

!N = N sin! i + N cos! j

m!g = !mg j

!fk = ! fk cos" i + fk sin" j .

So the vector sum of the forces on the block is

!FT =

!N + m!g +

!fk = (N sin! " fk cos!) i + (N cos! " mg + fk sin! ) j (1)

b) Now let’s choose unit vectors b pointing along the inclined plane and c pointing normal (perpendicular) to the plane as shown in the figure below.

With these unit vectors, the vector decomposition of the forces are

!N = N c

m!g = mg sin! b " mg cos! c

!fk = ! fkb .

So the vector sum of the forces on the block is

!FT =

!N + m!g +

!fk = (N ! mg cos" ) c + (mg sin" ! fk )b (2)

Suppose we want to find the magnitude of the acceleration of the block. Newton’s Second Law in the b -direction is then mg sin! " fk = mab (3) Because the block is constrained to move along the inclined plane ac = 0 , then Newton’s Second Law in the c -direction is then N ! mg cos" = 0 (4) The friction force law is fk = µk N = µk mg cos! (where we solved Eq. (4) for N ) so we

can solve Eq. (3) for the b -component of the acceleration ab = g(sin! " µk cos!) (5) which is also the magnitude of the acceleration. Suppose instead we were to us the i and j unit vectors and Eq. (1) for the vector decomposition of the sum of the forces.

Because the block is constrained to slide down the inclined plane, the component of the accelerations are constrained. In order to determine the constraint conditions we introduce a coordinate system as shown in the figure above. From the geometry we see that

tan! =

y(t)L " x(t)

.

Thus y(t) = (L ! x(t)) tan" (6) We can differentiate Eq. (6) twice and find that

ay = !ax tan" (7) In this coordinate system, using fk = µk N , Newton’s Second Law in the i -direction is then N (sin! " µk cos!) = max (8) Using fk = µk N and Eq. (7), Newton’s Second Law in the j -direction is then N (cos! + µk sin!) " mg = "max tan! (9) Let’s solve our equations for N , and ax . We first rewrite Eq. (9) as N (cos! + µk sin!) = mg " max tan! (10) Let’s start with N . We can multiply Eq. (8) by sin! yielding N (sin2! " µk sin! cos!) = max sin! (11) We can multiply Eq. (10) by cos! yielding N (cos2! + µk sin! cos!) = mg cos! " max tan! cos! = mg cos! " max sin! (12) we can now add Eqs. (11) and (12) using the trig identity cos2! + sin2! = 1 and solve for N : N = mg cos! (13) which is what we found already (Eq. (4)). We can now substitute Eq. (13) into Eq. (8) and find that

ax = g sin! cos! " gµk cos2! (14) We now use the trigonometric fact that a = ax / cos! in Eq. (14) to find the magnitude of the accelration

a =

ax

cos!= g(sin! " µk cos!) (15)

in agreement with our earlier result (Eq. (5)). We can now use Eq. (7) to find that

ay = !(g sin" cos" ! gµk cos2") tan"

which simplifies to

ay = !g sin2" + gµk sin" cos" (16)

This exercise should convince you to think carefully about how you choose your coordinate system.