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?
To find the magnitude of the minimum force of static friction required to hold both blocks at rest (a), we need to consider the forces acting on the blocks.
There are two main forces acting on the blocks: the weight of the suspended mass (73N) and the normal force exerted by the table on the 150N block. Since the blocks are at rest, the sum of the vertical forces must be zero.
Let's call the minimum force of static friction F_static. So the equation becomes:
73N - F_static - 150N = 0
Now we can solve for F_static:
F_static = 73N - 150N
F_static = -77N
It's important to note that the negative sign indicates that the direction of the force of static friction is opposite to the direction of motion. However, the magnitude of a force cannot be negative, so we take the absolute value:
Magnitude of F_static = |-77N|
Magnitude of F_static = 77N
Therefore, the magnitude of the minimum force of static friction required to hold both blocks at rest is 77N (answer for part a).
For part b, the minimum coefficient of static friction is the ratio of the force of static friction to the normal force between the block and the table.
The normal force is equal to the weight of the block resting on the table:
Normal force = 150N
So the minimum coefficient of static friction (μ_static_min) can be calculated as:
μ_static_min = F_static / Normal force
μ_static_min = 77N / 150N
μ_static_min = 0.5133
Therefore, the minimum coefficient of static friction required to ensure that both blocks remain at rest is approximately 0.5133 (answer for part b).
I hope this explanation helps! Let me know if you have any further questions.
To solve this problem, we need to analyze the forces acting on the blocks.
a) In this case, we have two blocks - one resting on the table with a weight of 150N and another suspended with a weight of 73N. The force of gravity acts vertically downward on each of these blocks.
Let's consider the forces acting on the block resting on the table.
- Force of gravity (weight): 150N (acting downwards)
- Normal force (reaction force from the table): The normal force will have the same magnitude as the weight because the block is at rest. Therefore, normal force = 150N (acting upwards)
- Static friction force: This force counteracts the tendency of the block to move. In this case, it prevents the block from sliding off the table.
Since the block is at rest, we can determine the magnitude of the minimum force of static friction by balancing the forces. The force of static friction must be equal in magnitude but opposite in direction to the force trying to slide the block off the table, which is the weight of the suspended mass (73N).
So the magnitude of the minimum force of static friction required to hold both blocks at rest is 73N.
b) The coefficient of static friction (μs) relates to the force of static friction. The equation for static friction is:
static friction = μs * normal force
We know the magnitude of the minimum force of static friction is 73N, and the normal force on the block resting on the table is 150N. Plugging these values into the equation, we get:
73N = μs * 150N
Now, we can solve for the minimum coefficient of static friction (μs):
μs = 73N / 150N
μs = 0.487
So, the minimum coefficient of static friction required to ensure that both blocks remain at rest is approximately 0.487.
In summary, the magnitude of the minimum force of static friction required to hold both blocks at rest is 73N, and the minimum coefficient of static friction required is approximately 0.487.