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Calculating Horizontal Force for a Shopping Cart on an Inclined Plane
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To calculate the horizontal force needed to push a 7.3 kg shopping cart up a 13° incline with an acceleration of 1.63 m/s², the formula \vec F = m \vec a is applicable. The gravitational force acting on the cart has a component parallel to the incline that must be considered. The horizontal force applied by the shopper also has a parallel component affecting the cart's motion. The sum of these forces parallel to the incline must equal the mass of the cart multiplied by the acceleration. Using these principles will lead to the correct calculation of the required horizontal force.
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Tide
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Use \vec F = m \vec a
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wr1015
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Tide said:
i did that and got 18.99 N but its being counted as wrong. There are some formulas in my book that show horizontal force as \sum\\F_{x}=F_{1},_{x}+F_{2},_{x}=F_{1}+F_{2}\cos\theta but am not given any forces??
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Tide
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You know what the force of gravity is! It's direction is vertical but has a component parallel to the incline. The shopper applies a horizontal force which also has a component parallel to the incline. The sum of the forces parallel to the incline equals mass times acceleration along the incline.
You should be able to take it from there.
You should be able to take it from there.
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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