Answer:
The value is 
Explanation:
From the question we are told that
The magnitude of the horizontal force is 
The mass of the crate is 
The acceleration of the crate is 
Generally the net force acting on the crate is mathematically represented as
Here 
is force of kinetic friction (in N) acting on the crate
So
=>
Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37
The answer to this would inFact be A
The magnitude of the force of friction is 40 N
Explanation:
To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:
- The constant pushing force forward, of magnitude F = 40 N
- The frictional force, acting backward,
Since the two forces are in opposite direction, the equation of motion is
where
m is the mass of the student+scooter
a is the acceleration
However, here the scooter is moving at constant speed: this means that its acceleration is zero, so
a = 0
And therefore,
which means that the magnitude of the force of friction is also equal to 40 N.
Learn more about force of friction here:
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Centripetal acceleration points from the object toward the center of the circular path it's traveling.
