A large box whose mass is 30 kg rests on a
frictionless floor. A mover pushes down on
the box with a force of 392 N at an angle 30◦
below the horizontal.
What is the acceleration of the box across
the floor?
Answer in units of m/s^2
The box now rests on a frictionless ramp angled at 15◦
. The mover pulls up on a rope
attached to the box to move it up the incline.
If the rope makes an angle of 25 ◦ with
the horizontal, what is the smallest force the
mover would have to exert to move the box
up the ramp? The acceleration due to gravity
is 9.81 m/s^2
Answer in units of N.
please someone help
force forward=392*cos30
acceleratin=forcforward/massbox
on the ramp section
force up the ramp=tension*cos5
force down the ramp due to gravity=massbox*g*sin25
net force up to move up >0 or
forceuptheramp>force downthe ramp sove for tension
To find the acceleration of the box across the floor, we can start by resolving the force into its horizontal and vertical components.
Given:
Force applied, F = 392 N
Angle below the horizontal, θ = 30°
Mass of the box, m = 30 kg
To find the horizontal component of the force, Fx:
Fx = F * cos(θ)
Fx = 392 N * cos(30°)
To find the vertical component of the force, Fy:
Fy = F * sin(θ)
Fy = 392 N * sin(30°)
Since the floor is frictionless, the net force acting horizontally will provide the acceleration:
Fnet = m * ax
ax = Fnet / m
The net horizontal force is equal to the horizontal component of the applied force:
Fnet = Fx
Plugging in the values:
ax = Fx / m
ax = (392 N * cos(30°)) / 30 kg
ax ≈ 6.779 m/s^2
Therefore, the acceleration of the box across the floor is approximately 6.779 m/s^2.
Now, to find the smallest force the mover would have to exert to move the box up the ramp, we can use similar steps.
Given:
Angle of the ramp, θ = 15°
Angle of the rope with the horizontal, α = 25°
Acceleration due to gravity, g = 9.81 m/s^2
To find the vertical component of the force due to gravity, Fg:
Fg = m * g
Fg = 30 kg * 9.81 m/s^2
To find the component of the applied force parallel to the incline, Fpar:
Fpar = F * cos(α)
Fpar = F * cos(25°)
Since the box is moving up the ramp, the net force acting parallel to the incline will provide the required acceleration:
Fnet = m * ag
ag = Fnet / m
The net force parallel to the incline is the difference between the parallel component of the applied force and the force due to gravity:
Fnet = Fpar - Fg
Plugging in the values:
ag = Fnet / m
ag = (F * cos(25°) - (30 kg * 9.81 m/s^2)) / 30 kg
The smallest force the mover would have to exert is when the acceleration is minimum, i.e., just enough to counteract the force due to gravity. Therefore:
ag = 0
So, we can solve for F:
0 = (F * cos(25°) - (30 kg * 9.81 m/s^2)) / 30 kg
Simplifying the equation:
F * cos(25°) = 30 kg * 9.81 m/s^2
F = (30 kg * 9.81 m/s^2) / cos(25°)
Therefore, the smallest force the mover would have to exert to move the box up the ramp is approximately (30 kg * 9.81 m/s^2) / cos(25°) N.
To find the acceleration of the box across the floor, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
Net force = mass * acceleration
In this case, the net force acting on the box is the horizontal component of the force applied by the mover. The horizontal component can be found using trigonometry:
Horizontal component = force * cos(angle)
where the angle is given as 30 degrees below the horizontal. Substituting the given values:
Horizontal component = 392 N * cos(30 degrees)
Next, we can divide the net force by the mass of the box to find the acceleration:
acceleration = (392 N * cos(30 degrees)) / 30 kg
Calculating this expression:
acceleration ≈ 6.77614 m/s^2 (rounded to 5 decimal places)
So, the acceleration of the box across the floor is approximately 6.77614 m/s^2.
Now, let's move on to the second part of the question to find the smallest force required to move the box up the ramp.
On the frictionless ramp, the force of gravity acting on the box is opposed by two forces: the force exerted by the mover and the component of the weight of the box acting parallel to the ramp. We only need to consider the forces parallel to the ramp for this calculation.
To determine the force required to move the box up the ramp, we need to balance the parallel component of gravity with the force applied by the mover:
Force applied by mover = parallel component of gravity
The parallel component of gravity can be calculated using trigonometry:
Parallel component of gravity = weight of the box * sin(angle)
where the angle is given as 15 degrees. Substituting the given values:
Parallel component of gravity = 30 kg * 9.81 m/s^2 * sin(15 degrees)
The force applied by the mover should be equal to or greater than the parallel component of gravity. So, the smallest force the mover would have to exert is:
Force applied by mover = 30 kg * 9.81 m/s^2 * sin(15 degrees)
Calculating this expression:
Force applied by mover ≈ 76.812 N (rounded to 3 decimal places)
Therefore, the smallest force the mover would have to exert to move the box up the ramp is approximately 76.812 N.