A 150N block rests on a table. The suspended mass has a weight of 73N.
a) What is the magnitude of the minimum force of static friction required to hold both blocks at rest? Answer in units of N
b)What minimum coefficient of static friction is required to ensure that both blocks remain at rest?
Can someone please explain the answer when they give it?
I am guessing these blocks are connected over a frictionless pulley.
hanging force=73 N.
that is pulling on the large block
a. 73=forcestaticfriction
b. 73=mu*mg solve for m u
What would you use for the m when you multiply it by 9.81?
To solve this problem, we need to consider the forces acting on the blocks.
a) The downward force acting on the 150N block is its weight, which is 150N. The upward force acting on this block is the force of static friction exerted by the table to keep the block at rest.
The suspended mass has a weight of 73N, which exerts a downward force on the 150N block. The force of static friction on the suspended mass is the force that prevents it from sliding down.
Since the blocks are at rest, the force of static friction must be equal to or greater than the force exerted by the 73N weight. Therefore, the minimum force of static friction required to hold both blocks at rest is 73N.
b) The coefficient of static friction is a measure of how rough or smooth the surfaces in contact are. It represents the ratio of the force of static friction to the normal force between the surfaces.
In this case, the normal force is the force exerted by the table on the 150N block, which is equal to its weight (150N).
To find the minimum coefficient of static friction required, we divide the force of static friction (73N) by the normal force (150N):
Coefficient of static friction = Force of static friction / Normal force
Coefficient of static friction = 73N / 150N
Coefficient of static friction ≈ 0.487
Therefore, a minimum coefficient of static friction of approximately 0.487 is required to ensure that both blocks remain at rest.
Note: The answer above assumes that both blocks have enough friction to keep them at rest. In some cases, if the blocks are on a very slippery surface, the coefficient of static friction required may exceed the maximum value of static friction between the surfaces. In such cases, additional measures may be needed to prevent the blocks from sliding.
To answer these questions, we need to understand the concept of static friction and how it relates to the forces acting on the blocks.
Static friction is the force that acts between two surfaces in contact when there is no relative motion between them. It always opposes the tendency of motion and prevents the objects from moving.
In this scenario, there are two forces acting on the blocks: the weight of the suspended mass (73N) pulling downward and the force of static friction pushing upward on the block resting on the table. These forces need to be balanced in order to keep the blocks at rest.
a) To find the magnitude of the minimum force of static friction required to hold both blocks at rest, we need to balance the forces. Since the block resting on the table is not moving, the force of static friction must be equal in magnitude and opposite in direction to the weight of the suspended mass (73N).
So, the magnitude of the minimum force of static friction required is 73N.
b) The coefficient of static friction (μs) is a dimensionless number that quantifies the relationship between the force of static friction and the normal force (the force exerted by a surface to support the weight of an object resting on it).
The equation to calculate the force of static friction is:
Force of static friction = coefficient of static friction × normal force
In this case, the normal force is the weight of the block resting on the table, which is equal to its mass multiplied by the acceleration due to gravity (150N).
To find the minimum coefficient of static friction required, we can rearrange the equation:
Coefficient of static friction = Force of static friction / Normal force
Substituting the values, we get:
Coefficient of static friction = 73N / 150N
Calculating this, the minimum coefficient of static friction required is approximately 0.487.
So, to summarize:
a) The magnitude of the minimum force of static friction required to hold both blocks at rest is 73N.
b) The minimum coefficient of static friction required is approximately 0.487.