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matt1687
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Homework Statement
A 6.2 kg block of ice is placed on a 17 degree ice incline. (u = 0.030)
Homework Equations
The Attempt at a Solution
I'm pretty much lost
Solving Forces on an Inclined Ice Block: 6.2 kg, 17 degrees, u = 0.030
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AI Thread Summary
A 6.2 kg block of ice is on a 17-degree incline with a coefficient of friction of 0.030. The problem requires creating a free body diagram (FBD) to visualize the forces acting on the block. Key forces include the gravitational force acting downward, the normal force perpendicular to the incline, and the frictional force opposing the motion. The goal is to calculate the net force acting on the block. Understanding these forces is essential for solving the problem effectively.
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
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