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Block slides inclined plane

  • Thread starter bigman8424
  • Start date
25
0
a 15.00 kb block slides down an included plane at 37.0 degrees to horizontal. find acceleration of block, if plane is frictionless:

theta: 37
m = 15.00 kg
a = ??


wtx - fk = ma
wtcos(53) - Fk = ma
15.00*9.8*.60 - Fk = ma
88 - Fk = ma
not sure here

n = mgcos37
n = 15.00*9.8*.80
n = 120 n

anyone no this 1
 
Last edited:

OlderDan

Science Advisor
Homework Helper
3,021
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bigman8424 said:
a 15.00 kb block slides down an included plane at 37.0 degrees to horizontal. find acceleration of block, if plane is frictionless:
find acceleration of block, if plane is frictionless:
 
391
2
You are only trying to find acceleration, bigman8424. You do not have to take weight into account because there is NO friction. What is the force of gravity going down the incline equal to on the block while it is sitting on the ramp?
 
651
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[tex] \sum_{along plane or wedge} F = mgsin37=ma [/tex]
 
25
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Ef = wt * sin(37) = ma
Ef = mgsin(37) = ma
(15.00)(9.8)(0.6) = (15.00)A
A = 5.9 m/s^2 ?
 
2,208
1
That's correct if you'll take the rounding error from sin(37) = 0.6.
 
25
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kinetic friction??

but, if the coefficient of kinetic friction was .20, then i would use my original formula:

(15.00)(9.8)(0.60) - Fk = MA
(15.00)(9.8)(0.60) - 0.20 = (15.00)A
 
2,208
1
The force of gravity pulls downwards at 9.8m/s^2. However, when an object acted upon by gravity is resting on another object, there is a normal force with a component that is countering the force of gravity. If the block was sitting on a table, then it is easy to see that as gravity pulls the block down, the table pushes it back up with exactly the opposite force (look at your monitor for example, its not moving). This is the case where [itex] \theta = 0 \ degrees [/tex].

If you held an object against a vertical frictionless wall and let go, the object would just slide down the wall at 9.8m/s^2 as there is no component of the normal force opposing gravity. Here [itex] \theta = 90 \ degrees [/itex],

For intermediate values, we can see that as theta goes from 0 to 90, the acceleration goes from 0 to 9.8, and the sin function gives us the proper scalar to show the effect of the opposing normal force on the net acceleration of the block.
 
651
1
yes agreed but Fk = 0.20 * 15 * g * cos37
 
2,208
1
bigman8424 said:
but, if the coefficient of kinetic friction was .20, then i would use my original formula:

(15.00)(9.8)(0.60) - Fk = MA
(15.00)(9.8)(0.60) - 0.20 = (15.00)A

The easiest way would be to draw a force diagram. Three forces, gravity, normal, and friction. Gravity pulls down at [itex] F_g = mg [/itex]. Normal pushes back up at [itex] F_n = mgcos(\theta) [/itex] perpendicular to the surface. The friction force is a scalar multiplied by the normal force, [itex] F_f = \mu mgcos(\theta) [/tex] along (upwards) the incline. The net force is the vector sum of all these forces. If you know your vectors, you can easily find the net force.
 
25
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sounds fun

is anyone good in elevator problems, i'm having trouble on a different problem, i need to find the acceleratin of a 600 n man on a bathroom scale in an elevator. once it starts moving, it reads 900 n
i posted the question on
https://www.physicsforums.com/showthread.php?t=79172
if anyone's interested
 
Last edited:

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