What does the atomic number of an atom tell us?
2 answers:
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This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip
Answer:
a) the total electric potential is 2282000 V
b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
Explanation:
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
so
Electric potential at p in the diagram 1 below is;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we know that; Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)
the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
r1² = 0.015² + 0.0125²
r1 = √[ 0.015² + 0.0125² ]
r1 = √0.00038125
r1 = 0.0195
Also
r2² = 0.015² + 0.018²
r2 = √[ 0.015² + 0.018² ]
r2 = √0.000549
r2 = 0.0234
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
