The diagram below shows a large cube of
mass 25 kg being accelerated across a frictionless
level floor by a horizontal force, F.
A small cube of mass 4.0 kg is in contact
with the front surface of the cube. The coefficient
of static friction between the cubes
is 0.71.
What is the minimum value of F such that the small cube will not slide down the large cube's side?
Start by drawing the fbd of the little box. Friction up, gravity down, normal towards the right. From that we can tell friction is the same as gravity.
Fg:
Fg=m•g so Fg=4•9.8
Fg=39.2
Ff=Fg so Ff=39.2
Fn:
Ff=μ•Fn so 39.2=.71•Fn
Fn=55.21126761
Calculate the acceleration:
ΣF=M•A
ΣFx=Fn=m•a
ΣFx=55.21126761=4•a
55.21126761/4=a
a=13.0828169
Then draw the fbd for the big box. Fn going up, left, and right and Fg going down. Calculate the y axis forces and the Fn going to the left.
Fg:
Fg=m•g so Fg=25•9.8
Fg=245 and Fn does as well
Fn:
Fn=55.21 because we solved for that on the small box. Bc of Newton's 3rd law we know those are paired forces.
ΣF=m•a
ΣFx=Fn-55.12=25•13.08
Fn-55.12= 345.0704225
Fn=400.28
The answer is 400.28
Well, F represents the force required to keep the small cube from sliding down the large cube. To find the minimum value of F, we need to consider the force of friction between the two cubes.
Now, the force of friction can be calculated by multiplying the coefficient of static friction (μs) by the normal force (Fn) between the two cubes. The normal force is equal to the weight of the small cube, which is given by m*g, where m is the mass of the small cube (4.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The normal force (Fn) = m * g = 4.0 kg * 9.8 m/s^2 = 39.2 N.
Now, let's calculate the force of friction (Ffriction).
Ffriction = μs * Fn = 0.71 * 39.2 N = 27.8 N.
So, the minimum value of F required to keep the small cube from sliding down the large cube is 27.8 N.
To find the minimum value of F such that the small cube will not slide down the large cube's side, we need to consider the forces acting on the small cube.
There are two main forces acting on the small cube. The first force is the gravitational force, which can be calculated using the formula:
F_gravity = m * g
Where m is the mass of the small cube and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The second force is the static friction force between the cubes, which can be calculated using the formula:
F_friction = μ * N
Where μ is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the small cube, which is given by:
N = m * g
Therefore, the maximum static friction force is:
F_max = μ * m * g
Since the small cube will not slide down the large cube's side, the maximum static friction force must be equal to the gravitational force. Therefore:
F_max = F_gravity
μ * m * g = m * g
Simplifying the equation, we can cancel out m * g on both sides:
μ = 1
Therefore, the minimum value of F is equal to the gravitational force acting on the small cube:
F = m * g
Substituting the values given in the problem, we have:
F = 4.0 kg * 9.8 m/s^2
F = 39.2 N
So, the minimum value of F such that the small cube will not slide down the large cube's side is 39.2 N.
To find the minimum value of F such that the small cube will not slide down the large cube's side, we need to consider the forces acting on the small cube.
1. Identify the forces acting vertically on the small cube:
- The weight of the small cube (mg), acting downward.
- The normal force (N), exerted by the large cube on the small cube, acting upward.
2. Determine the coefficient of static friction between the cubes (μs). In this case, it is given as 0.71.
3. Recognize that the maximum static friction force (fs_max) between the cubes is given by:
fs_max = μs * N
4. Consider the forces acting horizontally on the small cube:
- The force of friction (f) between the large and small cubes.
- The applied force (F), exerted on the large cube.
Since the frictional force and the applied force are the only horizontal forces, the small cube will not slide down if the frictional force can balance the component of the weight of the small cube that is acting parallel to the surface of the large cube.
5. Determine the component of the weight of the small cube acting parallel to the surface of the large cube:
- The weight (mg) can be split into two components: one parallel to the surface (mg sinθ) and one perpendicular to the surface (mg cosθ).
- Since the large and small cubes are in contact, the angle θ between the surface of the large cube and the vertical direction is the same as theta between the weight and the vertical direction.
6. Equate the force of friction to the component of the weight acting parallel to the surface:
f = mg * sinθ
7. Substitute the maximum static friction force into the equation:
μs * N = mg * sinθ
8. Solve for the normal force (N):
N = mg * sinθ / μs
9. Substitute the normal force into the maximum static friction force equation:
fs_max = μs * (mg * sinθ / μs) = mg * sinθ
10. Finally, equate the force of friction to the applied force:
f = F
11. Substitute the component of the weight and the maximum static friction force into the equation:
mg * sinθ = F
Therefore, the minimum value of F such that the small cube will not slide down the large cube's side is mg * sinθ.
Please note that to calculate the exact value, you will need to know the angle θ between the surface of the large cube and the vertical direction.