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Gennadij [26K]
3 years ago
15

Help with question 19 please

Physics
1 answer:
juin [17]3 years ago
6 0

Answer:

4.2m/s

Explanation:

Now the velocity or speed initiated by the 200g mass at 7.0m/s is related by the formula;

V=ω×r ; ω is the angular velocity

r is the distance

Note that;

ω = √k/m. ; where K is the Force constant of the spring.

m is the mass

We see from the two expressions that v will only change with r if the spring constant and mass stays the same.

Hence we can rewrite the equation of v for two pairs of velocity V1 , V2 and distance r1 , r2 as;

V1/r1 = V2/R2

If V1=7.0m/s , r1=25cm= 0.25m

V2=? r2 = 15cm = 0.15m

Hence we can calculate V2 and is;

V2 = V1 × R2/ R1

V2 = 7 × 0.15 / 0.25= 4.2m/s

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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 fricti
Colt1911 [192]

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

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Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

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FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

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NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

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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

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