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FREE BODY DIAGRAMS!For each of the layouts, draw the f.b.d.s for the bodies in the system. (The “solutions” follow--try each before looking!) !
1.)!
1.)!
3m
frictionless!
2.)!
3m
m
2.)!
frictional!
v!3.)!
3m
m
3.)!
frictional!
v!4.)!
3m
m
4.)!
3m1.) answer
3m( )g
N
5.)!
frictionless!
3m
m
2.) answer
T
T
mhg3m( )g
N
6.)!
frictional!
v!3.) answer !3m( )
m
T
mhg3m( )g
N
fk = µkN
T
7.)!
frictional!
v!4.) answer!
3m
m
T
mhg3m( )g
N
fk = µkN
T
8.)!
5.)! frictionless!3m
v
θ
9.)!
6.)! frictional!3m
v
θ
10.)!
7.!
frictional!
3mm
θ
11.)!
8.)!
3m
2m
12.)!
5.) answer!
frictionless!3m
v
N
3m( )gθ
θ
13.)!
6.) answer!
frictional!
3m
v
θ
N
3m( )gθ
fk = µkN
14.)!
7.) answer!
frictional!
3mm
θ
N
3m( )g θ
fk = µkN
T
mg
T
15.)!
8.) answer!
3m
2m T T
3m( )g 2m( )g
16.)!
9.)!
3m
2m
17.)!
7m
2m
m
10.)!
frictional on both surfaces; (7m mass moving to left initially with 2m mass just barely holding on)!
v
18.)!
7m
2m
m
11.)!
frictional on both surfaces; (7m mass moving to right initially with 2m mass breaking loose)!
v
19.)!
12.)!
7m
2m
m
frictional on both surfaces and assume everyone has just broken loose and is accelerating in its appropriate direction (that is, there’s no static friction but there is kinetic friction).!
(Note that because there are no external force in the system in the horizontal, the system’s center of mass must stay where it is. That means any motion of the 7m mass must be countered by motion of the 2m mass IN THE OPPOSITE DIRECTION.) If this seems too obscure to wrap your brain around, just take a quick look at the solution and go on. Don’t take a lot of time on this!!
20.)!
9.) answer!
3m
2mT T
3m( )g 2m( )g
21.)!
10.) answer! frictional on both surfaces; (7m mass moving to left initially with 2m mass just barely holding on)!
2m( )g
7m( )g
mg
N1
N1
Nfloor
T
T
µfloor,kineticNfloor
µ1,staticN1
µ1,staticN1
22.)!
Comment: Notice that gravity acting on the hanging mass is the only external force in the system (disconnect “m” and nothing accelerates). That means it governs the net acceleration. As the hanging mass is accelerating downward, both the 7m and 2m masses must be accelerating to the right (even though they are moving to the left), which means they are slowing down and there better be a net force to the right in both cases. The only place the 2m guy can get its right-directed force is from static friction between it and the 7m mass. That means there’s a static frictional force on the 7m mass to the left (Newton’s Third Law). Also, there’s a kinetic frictional force on the 7m mass due to the floor and two different normal forces to deal with. Fun, eh?!
7m
2m
m
v
7m
2m
m
11.) answer!frictional on both surfaces; (7m mass moving to right initially with 2m mass breaking loose)!
v
2m( )g
7m( )g
mg
N1
N1 Nfloor
T
T
µfloorNfloor
µ1,kineticN1
µ1,kineticN1
23.)!
Comment: As in #10, gravity on the hanging mass governs, so the hanging mass must accelerate downward and the 7m mass must accelerate (speeding up) to the right. Even though the 2m mass breaks loose, it will get dragged along to the right by its interaction with the 7m mass. As the only force acting in the horizontal on the 2m mass is kinetic friction, that force must be to the right. That means a kinetic frictional force must also act on the 7m mass to the left (Newton’s Third) diminish the magnitude of the 7m mass’s acceleration to the right. !
12.) answer!
7m
2m
m
Comment: As before, the hanging mass accelerates downward making, in this case, the 2m mass accelerate to the right. There will be kinetic friction between the 2m and 7m mass with that frictional force retarding 2m’s acceleration by being directed to the left. By the same token, the 2m mass will try to drag the 7m mass to the right via that same kinetic frictional force (NTL). In other words, you might think that both the 2m and 7m blocks would accelerate to the right. The problem is that the center of mass of this system has to stay in place (there is no outside force acting in the horizontal, so the center of mass stays put). The only way that can happen is if the 7m mass accelerates to the left. But how could it do that? It does it due to the tension force in the line that essentially acts on the 7m mass to the left (look at the sketch). If the pulley was attached to the table, this wouldn’t be the case, but because it is attached to the 7m mass, you have to include that tension force as it act both in the horizontal and vertical on the 7m mass! Even more fun!!
2m( )g
7m( )g
mg
N1
N1 Nfloor
T
µfloorNfloor
µ1,kineticN1
T
µ1,kineticN1
T
T
24.)!
frictional on both surfaces with velocities shown (this is tricky): !
13.)!frictional on both surfaces; (velocities of 5m and 3m masses are down incline--acceleration assumed UP the incline with 3m mass breaking loose)!
3m
m5m
θ
v1
25.)!
v2
3m
m5m
θ
14.)!
frictional on both surfaces; (velocities shown with 3m mass breaking loose and both accelerating DOWN incline, which means they are slowing down)!
v1
26.)!
v2
13.) answer! 3m
m5m
θ
v1
3m( )g
5m( )g
mg
N1
N1
Nfloor
T
µfloorNfloor
µ1,kineticN1
µ1,kineticN1
T
frictional on both surfaces; (velocities of 5m and 3m masses are down incline--acceleration assumed UP the incline and 3m mass has broken loose)!
27.)!
Comment: Don’t take a lot of time on this as it is tricky. If both the 3m and 5m blocks are moving down the incline, and if both are slowing (as must be the case if the acceleration is UP the incline), how would the unencumbered 3m mass act relative to the 5m mass with a tension pulling back on the 5m mass? The 5m mass is being slowed down by the tension force on it, whereas the 3m mass will slow down ONLY due to friction between it and the 5m mass. That means the kinetic friction force on the 3m mass must be Up the incline. Due to NTL, a “reaction” kinetic friction force must be exerted on the 5m mass down the incline. There will also be a kinetic friction force due to the table on the 5m mass that is up the incline (the 5m guy is moving down the incline), and there are two normal forces to deal with.!
v2
3m
m5m
θ
14.) answer! frictional on both surfaces; (velocity of 5m mass shown with 3m mass breaking loose and both accelerating DOWN incline)!
3m( )g
5m( )g
mg
N1
N1
Nfloor
T
µfloorNfloor
µ1,kineticN1
µ1,kineticN1
T
28.)!
v1
v2
Comment: Don’t take a lot of time on this as it is tricky. If both the 3m and 5m blocks are moving up the incline, and if both are slowing (as must be the case if the acceleration is DOWN the incline), what is motivating the masses to slow? It’s the component of gravity down the incline. The component of the 5m mass will be larger than the component of the 3m mass, so the 5m mass will slow down faster (the 3m mass also has a tension force pulling it UP the incline). That means the 5m mass will be moving slower than the 3m mass and the 3m mass will be moving toward the top of the incline RELATIVE TO THE 5M MASS! In other words, the kinetic frictional force on the 3m mass will be DOWN the incline. NTL maintains that there must, therefore, be a kinetic friction force on the 5m mass due to the 3m mass that is UP the incline (see f.b.d.).!
