Force & Motion: Find Max Weight of Block A for Stationary System

[Solms]

Hi, iam new here and would like some help with my high school physics. i would like to know the steps (and explanations), not just the answer. since i need to do good on the tests.

Block B weighs 711 N. the coefficient of static frcition between block and the table is 0.25; assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be staionary.
.../|(wall)
...../..|
.../...|
(B)----o(30degrees)
____...|
table|..|
...(A)
(ignore the periods)

 

[Päällikkö]

You are supposed to show what you've tried:
https://www.physicsforums.com/showthread.php?t=28

Draw a free body diagram.
What is the force pulling B off the table? How is this force affected by A's weight?

 

[quicknote]

Hello and welcome to the world of physics! I am happy to assist you with your question about force and motion. Let's break down the problem step by step.

First, we need to understand the concept of force and motion. Force is a push or pull on an object that causes it to accelerate or change its state of motion. Motion, on the other hand, is the movement of an object from one place to another.

In this problem, we have two blocks, A and B, connected by a cord. Block B has a weight of 711 N, which means that it is being pulled down towards the ground with a force of 711 N. Our goal is to find the maximum weight of block A that will keep the system stationary, meaning that there will be no movement in any direction.

To solve this problem, we need to consider the forces acting on the system. There are two main forces at play here - the weight of block B pulling down and the force of static friction between block B and the table.

The coefficient of static friction, which is given as 0.25, tells us the maximum amount of friction that can exist between two surfaces before they start to slide against each other. In this case, it means that the table can apply a maximum frictional force of 0.25 times the weight of block B, which is 0.25 x 711 N = 177.75 N.

Now, let's look at the forces acting on block A. The only force acting on it is the tension in the cord, which is pulling it towards block B. Since the system is stationary, the tension in the cord must be equal to the maximum frictional force, which is 177.75 N.

We can now use Newton's Second Law of Motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the net force on block A is the tension in the cord, and its mass is the unknown weight we are trying to find. We can set up the following equation:

Tension = Mass x Acceleration

177.75 N = Mass of A x 0 m/s^2

Since the acceleration is zero (the system is stationary), the mass of block A must also be zero. This means that block A cannot have any weight for the system to be stationary.

In conclusion, the maximum weight of block A for the system to be stationary is zero.