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djverab [1.8K]
3 years ago
12

2 Blocks 1 and 2 rest on rough surfaces with coefficient of frictions ¢1 and ¢2 respectively. The blocks

Engineering
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Answer:

  • 100N
  • 25N

Explanation:

a) On the verge of tipping over, reaction acts at the corner A

When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at A must pass through B

tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )

b) When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at C must pass through G

k tanθ µ =

tan x/ H/2 , x H/2

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Explanation:

https://opendsa-server.cs.vt.edu/embed/mergesortAV

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Please help me it’s for science I only have a few minutes
7nadin3 [17]

Answer:

Rocks

Explanation:

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2 years ago
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When your complex reaction time is compromised by alcohol, an impaired person's ability to respond to emergency or unanticipated
babunello [35]

Answer:

decreased

Explanation:

when impaired you react slower then you would sober.

3 0
3 years ago
A 14-lb crate is pulled up a frictionless 40° ramp with an initial velocity of v1=0.4 ft/s. It is pulled 0.3 ft from location #1
Morgarella [4.7K]

Answer:

3.25 ft/s

Explanation:

The crate is of =14-lb=m₁

The angle of inclination is = 40°=Ф

The initial velocity = 0.4 ft/s= v₁

Distance the crate will move is= 0.3 ft =d

The load pulling downwards is = 36 lb= m₂

Acceleration of the pulley, a= m₂g - m₁gsinФ / m₁+m₂ where g= 32.17 ft/s^2

a= 36*32.17 - 14*32.17*sin 40° / 14+36

a=17.37 ft/s^2

Apply the formula for final velocity

V₂²=V₁²+2ad

V₂²=0.4²+ 2*17.37*0.3

V₂²=10.582

V₂ =√10.582 = 3.25 ft/s

6 0
3 years ago
I need help with this question please
solniwko [45]

Answer:

The resultant moment is 477.84 N·m

Explanation:

We note that the resultant moment is given by the moment about a given point

The length of the sides of the formed triangles are;

l = sin(40°) × 4/sin(110°) ≈ 2.736

Taking the moment about the lower left hand corner of the figure, with the convention that clockwise moments are positive, we have;

The resultant moment, ∑m, is given as follow;

∑M = 250 N × 4 m + 400 N × cos(40°) × 4 m - 400 N × cos(40°) × 2 m + 400 N × sin(40°) × 2 m × tan(40°) - 600 N × cos(40°) × 2 m - 600 N× sin(40°) × 2 m × tan(40°) = 477.837084 N·m

Therefore, the resultant moment, ∑m ≈ 477.84 N·m clockwise.

6 0
3 years ago
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