A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle
of 40 degrees to the ramp, how hard would you have to push to get the box to move up the ramp?
1 answer:
<u>Given data:</u>
m= 50 Kg,
W= m×g = 50 × 9.81 = 490.5 N
ramp angle (α) = 30 degrees,
coefficient of friction (μs) = 0.5773,
Push at an angle (Θ) = 40 degrees,
Determine: Push to get box move up (P)=?
From the figure,
Resolving the forces along the plane
W sinα + μs.R = P cos Θ --------------------- (i)
Resolving the forces perpendicular to the inclined plane
W cosα = R+Psin Θ => R= W cosα - Psin Θ -------------- (ii)
Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>
<em>Pmin = W sin( α +Θ )</em>
<em> = W[ sin α.Cos Θ + cos α.sin Θ]</em>
<em> = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>
<em> = 460.9 N</em>
<em>Minimum push required to move the box up the ramp is 460.9 N</em>
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Answer:
Explanation:
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where <em>f</em> is the frequency, <em>v</em> is the velocity/wave speed, and λ is the wavelength.
The wavelength is 10 meters and the velocity is 200 meters per second.
- 1 m/s can also be written as 1 m*s^-1
Therefore:
Substitute the values into the formula.
Divide and note that the meters (m) will cancel each other out.
- 1 s^-1 is equal to Hertz
- Therefore, our answer of 20 s^-1 is equal to 20 Hz
The frequency of the wave is <u>20 Hertz</u>
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Explanation:
The electric field is defined as the change in the properties of space caused by the existence of a positively (+) or negatively (-) charged particle. The electric field can be represented by infinitely many lines from a particle, and those lines never intersect each other. Depending on the type of charge we can see different cases:
- Let's say we have a <u>positive charge alone (</u>image 1)<u>.</u> The field lines are drawn from the centre of the particle outwards to infinity (in other words, they disappear from the edge of the picture). Meaning the direction of the electric field points outwards the particle.
- For a <u>negative charge alone </u>(image 2)<u>,</u> the lines come from infinity to the centre, and point towards the particle (i.e. lines appear from the edge of the picture).
Let's see what happens if we have two charges together:
- <u>Two positive charges</u> (image 3): Since the charges are of the same type (positive), the particles repel each other. Then the field lines will avoid each other so they do not join. The charge is positive, so lines point outwards.
- <u>Two negative charges</u> (image 4): Again, the charges are both negative, so they repel. But they are negative, so the field points inwards.
- <u>Negative and positive charges</u> (image 5): They are different charges, so the force between them is attractive. This causes the field lines from both to join. They go out of the positive and come into the negative particle.
Image 6:
The lines are passing through infinite points of the space. If we choose a certain point and measure the electric field, we can see to which direction the electric field points. This is the direction of the electric field vector. It does not matter which point we choose; the electric field vector touches the field line only at this point, which means it is tangent to the field line.
