Physics 206 Example Problems Newton's Laws of Motion

Solution :
There is a range of masses because static friction provides, up to its maximum value, the force it needs to
maintain equilibrium.
Maximum m1 : The maximum value of m1 is obtained by letting static friction take on its maximum
value and act down the incline. For m1 I’ll use a coordinate system such that up is +y. For m2
I’ll let + x be parallel to and down the incline and +y be in the direction of the outward
normal to the incline. The free body diagrams for the masses look like:
In the chosen coordinate systems:
Fron1
m1
FG1
Fron1
= Tyˆ FG1
= − m1 gE yˆ
Fron2 = − Txˆ FG2
= m2 gE sinθ xˆ − m2 gE cosθ yˆ N = Nyˆ FS
= µS Nxˆ
Newton’ s second law applied in the y-direction on mass 1 gives,
Fron2
m2
θ
FG2
T − m1 gE = m1 a1 y = 0 ⇒ T = m1 gE
Newton’ s second law applied in the y-direction on block 2 gives,
Newton’ s second law applied in the x-direction on mass 2 gives,
Using the results of Equations 1 and 2 in Equation 3 gives,
Solving for m1 gives,
N
FS
y
x
N − m2 gE cosθ = 0 ⇒ N = m2 gE cosθ ( 2)
µS N + m2 gE sinθ − T = 0 ( 3)
µS m2 gE cosθ + m2 gE sinθ − m1 gE = 0
m1 = µs m2 cosθ + m2 sinθ = { ( 0. 4) ( 1 0) cos( 30 ◦ ) + ( 1 0) sin( 30 ◦ ) } kg ( 4)
m1 , m ax = 8. 46 kg
Minimum m1 : The minimum value of m1 is obtained by setting the static frictional force to its max-
imum value but letting it act up the incline so that F
S = − µS Nxˆ . The development leading to Equation
4 still holds except that a negative sign precedes the term involving µS. That is,
m1 = − µs m2 cosθ + m2 sinθ = { − ( 0. 4) ( 1 0) cos( 30 ◦ ) + ( 1 0) sin( 30 ◦ ) } kg ( 5)
m1 , m ax = 1 . 54 kg
The hanging block may have any mass between 1 . 54 kg and 8. 64 kg and the system will be in static equilibrium.
B) Equation ( 1 ) gives the range of tensions as 1 5N to 83N.
Problem 8. The coefficient of static friction between a box of mass 1 0 kg and a truck bed is 0. 4. What
is the maximum possible acceleration of the truck for which the box does not slide on the truck bed?
5
( 1 )