- #1
bumblebeeliz
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Homework Statement
A 4.0-kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on top of the
toboggan. The coefficient of static friction µs between the block and the surface of the toboggan is 0.60, whereas the kinetic friction coefficient is 0.51. The block is pulled by a
30 N horizontal force as shown. What are the magnitudes and directions of the
resulting accelerations of the block and the toboggan?
Homework Equations
Ffr = [tex]\mu[/tex]s Fn = [tex]\mu[/tex]s mg
Ffr = [tex]\mu[/tex]k Fn = [tex]\mu[/tex]k mg
Fn = mg
[tex]\Sigma[/tex]F= ma
The Attempt at a Solution
Fn= (m1 + m2) g
Fn= (2.0kg + 4.0kg) (9.80m/s2)
Fn= 58.8 N
Now I calculate the friction of both separately: (im not sure about this part)
Box:
Ffr = 0.60 (58.8N)
Ffr = 35.25N
Tob:
Ffr = 0.51 (58.8N)
Ffr= 29.988N
Then I attempt the acceleration:
[tex]\Sigma[/tex]=m1a
Fr-F=ma
35.25N-30N = (2kg) a
5.25N/2kg = a
2.625 m/s2 = a
[tex]\Sigma[/tex]= m2a
F-Ffr =ma
30N-29.988N = (4kg) a
0.003 m/s2 = a (or 3 x 10-3)
This is where I get confuse. When the box is pulled doesn't it stay static while the sled slides on the ice? And what do they mean by direction?
Any help would be great :)