First, we will get the resultant force:
The direction of the force due to the person's weight is vertically down.
weight of person = 700 newton
Assume that the force exerted by the arms has a vertically upwards direction.
Force exerted by arms = 2*355 = 710 newtons
Therefore, the resultant force = 710 - 700 = 10 newtons (in the vertically upwards direction)
Now, we will get the mass of the person.
weight = 700 newtons
weight = mass * acceleration due to gravity
700 = 9.8*mass
mass = 71.428 kg
Then we will calculate the acceleration of the resultant force:
Force = mass*acceleration
10 = 71.428*acceleration
acceleration = 0.14 m/sec^2
Finally, we will use the equation of motion to get the final speed of the person.
V^2 = U^2 + 2aS where:
V is the final velocity that we need to calculate
U is the initial velocity = 0 m/sec (person starts at rest)
a is the person's acceleration = 0.14 m/sec^2
S is the distance covered = 25 cm = 0.25 meters
Substitute with the givens in the above equation to get the final speed as follows:
V^2 = U^2 + 2aS
V^2 = (0)^2 + 2(0.14)(0.25)
V^2 = 0.07
V = 0.2645 m/sec
Based on the above calculations:
The person's speed at the given point is 0.2645 m/sec
Hooke's Law says that F=-kx where k is the spring constant measured in N/m (newtons per meter)
Answer:
this description is valid for mediadle displacement, bone is an acceptable description
Explanation:
The description of a person's position must be done with a position vector. These vectors must have magnitude, a given direction and a starting point.
In the description this has a starting point corner NO of pine and 675.
Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.
After analyzing this description is valid for mediadle displacement, bone is an acceptable description
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
Answer:
Im pretty sure its number 2
Explanation:
