The drawing shows a large cube (mass = 27 kg) being accelerated across a horizontal frictionless surface by a horizontal force vector P . A small cube (mass = 4.0 kg) is in contact with the front surface of the large cube and will slide downward unless vector P is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that vector P can have in order to keep the small cube from sliding downward?
chuck norris
how does a small cube go "downward" if both are on a horizontal surfce? I don't understand the picture painted.
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That is what the picture looks like. Hopefully the formatting won't mess up once it is posted. Its like one cube is temporarily glued to the other.
Since the formatting messed up...
Imagine a regular, large cube. A smaller cube is placed on the right hand side of it but it appears as if it is glued to the side of the cube a couple inches above the horizontal surface.
ok, I get it. force on the surface keeps it up (actually friction does.)
The normal force on the front surface of the large cube to the small cube is
forcenormal=smallcubemass*a
the friction force then holding up the small cube is forcenormal*mu
and that has to equal smallcubemass*g
so smallcubemass*g=smallcubemass*a*mu
or mu=a/g
Now, what is a? Pushing force= (total mass)a or a=P/totalmass
so we have...
mu=a/g=P/(totalmass*g)
solve for P.
thanks...got the right answer!
To find the smallest magnitude of vector P required to keep the small cube from sliding downward, we need to analyze the forces acting on both cubes.
Let's start by considering the forces acting on the small cube:
1. Weight (mg): The small cube experiences a downward force due to its weight, which is given by the mass (m) of the small cube multiplied by the acceleration due to gravity (g). So, the weight of the small cube is W = (4.0 kg) x (9.8 m/s^2) = 39.2 N.
2. Friction force (f): The small cube experiences a friction force opposing its motion, which can be calculated using the coefficient of static friction (μs). The maximum static friction force that can act between the cubes is given by fs = μs * N, where N is the normal force. The normal force is the force exerted on the small cube by the large cube, which is equal to its weight (N = mg). So, the maximum static friction force is fs = (0.71) * (4.0 kg) * (9.8 m/s^2) = 27.976 N.
Now, let's analyze the forces acting on the large cube:
1. Weight (Mg): The large cube experiences a downward force due to its weight, which is given by the mass (M) of the large cube multiplied by the acceleration due to gravity (g). So, the weight of the large cube is W = (27 kg) x (9.8 m/s^2) = 264.6 N.
2. Force P (P): We want to find the minimum magnitude of vector P required to prevent the small cube from sliding downward. This force exerts an upward force on the small cube through the contact surface, preventing it from sliding. Since the small cube is in contact with the front surface of the large cube, the force P also acts on the large cube in the opposite direction.
Now, if vector P is large enough, it will prevent the small cube from sliding downward. In this case, the static friction force (fs) acting on the small cube will be equal to the weight (W) of the small cube.
Therefore, to keep the small cube from sliding downward, vector P must be equal to or greater than the maximum static friction force acting on the small cube, which in this case is fs = 27.976 N.