The drawing shows a large cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force . A small cube (mass = 4.0 kg) is in contact with the front surface of the large cube and will slide downward unless is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that can have in order to keep the small cube from sliding downward?
first we find the normal force on the small cube:
F=ma (where f=force,a=acceleration,m=mass)
a=f/m
acceleration on small box is same as that on the large box. let force to find be P.
then:
a=p/(25kg+4kg)
a=p/(29kg)m/s^2
force acting on small box:
f=ma
f=4*(p/29)N(normal force)
friction force= Us*(normal force) Us is coeff. of static friction.
friction= 0.71*(4*p/29)
weight= mg= 4*9.8
for the object(small box) not to slide down the friction force b/w the two objects have to be exactly the same as the weight of the object.
0.71*(4*p/29)=4*9.8
solving..
p= 400.29N
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
will there be Ff as reaction force going vertically down on the big box?
Well, it seems like we've got a slippery situation here! Let's calculate the smallest magnitude of the force required to prevent the small cube from sliding downward.
First, we need to determine the maximum static friction force between the cubes. Using the coefficient of static friction (μs = 0.71) and the normal force between the cubes, we can calculate the maximum static friction force (fsmax).
The normal force is the weight of the small cube, which is given by the formula N = 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).
So, N = 4.0 kg * 9.8 m/s^2 = 39.2 N.
Now, we can find the maximum static friction force by multiplying the coefficient of static friction (0.71) by the normal force (N).
fsmax = μs * N = 0.71 * 39.2 N = 27.83 N.
To prevent the small cube from sliding downward, the force applied to the large cube must be greater than or equal to the maximum static friction force. Therefore, F ≥ fsmax.
Hence, the smallest magnitude that the force can have is 27.83 N. Anything less than that, and the little cube will go on a slippery slide downward!
To find the smallest magnitude of force required to keep the small cube from sliding downward, we first need to determine the maximum friction force between the cubes.
The maximum static friction force can be calculated using the formula:
\( F_{\text{{friction}}} = \mu_s \cdot F_{\text{{normal}}} \)
where
\( \mu_s \) is the coefficient of static friction,
\( F_{\text{{normal}}} \) is the normal force between the two cubes.
Since the small cube is in contact with the large cube, the normal force between them is equal to the weight of the small cube, which is given by:
\( F_{\text{{normal}}} = m_{\text{{small}}} \cdot g \)
where
\( m_{\text{{small}}} \) is the mass of the small cube,
\( g \) is the acceleration due to gravity (approximately 9.8 m/s²).
Now, we can substitute the values into the formula to find the maximum static friction force:
\( F_{\text{{friction}}} = \mu_s \cdot m_{\text{{small}}} \cdot g \)
\( F_{\text{{friction}}} = 0.71 \cdot 4.0 \ \text{{kg}} \cdot 9.8 \ \text{{m/s²}} \)
\( F_{\text{{friction}}} \approx 27.68 \ \text{{N}} \)
To keep the small cube from sliding downward, the force applied must be equal to or greater than the maximum static friction force between the cubes. Therefore, the smallest magnitude that the force can have is 27.68 N.