Physics 206 Example Problems Newton's Laws of Motion
Physics 206 Example Problems Newton's Laws of Motion
Physics 206 Example Problems Newton's Laws of Motion
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Solution :<br />
There is a range <strong>of</strong> masses because static friction provides, up to its maximum value, the force it needs to<br />
maintain equilibrium.<br />
Maximum m1 : The maximum value <strong>of</strong> m1 is obtained by letting static friction take on its maximum<br />
value and act down the incline. For m1 I’ll use a coordinate system such that up is +y. For m2<br />
I’ll let + x be parallel to and down the incline and +y be in the direction <strong>of</strong> the outward<br />
normal to the incline. The free body diagrams for the masses look like:<br />
In the chosen coordinate systems:<br />
Fron1<br />
m1<br />
FG1<br />
Fron1<br />
= Tyˆ FG1<br />
= − m1 gE yˆ<br />
Fron2 = − Txˆ FG2<br />
= m2 gE sinθ xˆ − m2 gE cosθ yˆ N = Nyˆ FS<br />
= µS Nxˆ<br />
Newton’ s second law applied in the y-direction on mass 1 gives,<br />
Fron2<br />
m2<br />
θ<br />
FG2<br />
T − m1 gE = m1 a1 y = 0 ⇒ T = m1 gE<br />
Newton’ s second law applied in the y-direction on block 2 gives,<br />
Newton’ s second law applied in the x-direction on mass 2 gives,<br />
Using the results <strong>of</strong> Equations 1 and 2 in Equation 3 gives,<br />
Solving for m1 gives,<br />
N<br />
FS<br />
y<br />
x<br />
N − m2 gE cosθ = 0 ⇒ N = m2 gE cosθ ( 2)<br />
µS N + m2 gE sinθ − T = 0 ( 3)<br />
µS m2 gE cosθ + m2 gE sinθ − m1 gE = 0<br />
m1 = µs m2 cosθ + m2 sinθ = { ( 0. 4) ( 1 0) cos( 30 ◦ ) + ( 1 0) sin( 30 ◦ ) } kg ( 4)<br />
m1 , m ax = 8. 46 kg<br />
Minimum m1 : The minimum value <strong>of</strong> m1 is obtained by setting the static frictional force to its max-<br />
imum value but letting it act up the incline so that F<br />
S = − µS Nxˆ . The development leading to Equation<br />
4 still holds except that a negative sign precedes the term involving µS. That is,<br />
m1 = − µs m2 cosθ + m2 sinθ = { − ( 0. 4) ( 1 0) cos( 30 ◦ ) + ( 1 0) sin( 30 ◦ ) } kg ( 5)<br />
m1 , m ax = 1 . 54 kg<br />
The hanging block may have any mass between 1 . 54 kg and 8. 64 kg and the system will be in static equilibrium.<br />
B) Equation ( 1 ) gives the range <strong>of</strong> tensions as 1 5N to 83N.<br />
Problem 8. The coefficient <strong>of</strong> static friction between a box <strong>of</strong> mass 1 0 kg and a truck bed is 0. 4. What<br />
is the maximum possible acceleration <strong>of</strong> the truck for which the box does not slide on the truck bed?<br />
5<br />
( 1 )