In the figure, blocks A and B have weights of 49 N and 32 N, respectively. Determine the minimum weight of block C to keep A from sliding, if the coefficient of static friction between A and the table is 0.20.
There must be a diagram for this, I have no idea where a, B, and C are.
To determine the minimum weight of block C to keep block A from sliding, we need to consider the forces involved and apply Newton's second law of motion.
First, let's consider the forces acting on block A. There are two main forces: the gravitational force (weight) acting downwards and the static friction force acting upwards. The weight of block A, given as 49 N, acts straight down. The static friction force opposes the tendency of block A to slide and acts perpendicular to the contact surface between block A and the table.
The static friction force can be calculated using the formula:
F_friction = coefficient of static friction * Normal force
Here, the coefficient of static friction is given as 0.20, but we need to determine the normal force between block A and the table.
The normal force is equal to the weight of block A since the block is not accelerating vertically. Therefore, the normal force is also 49 N.
Now, we can calculate the static friction force:
F_friction = 0.20 * 49 N = 9.8 N
The static friction force acts in the opposite direction to the applied force, which is the weight of block C acting down.
Therefore, to prevent block A from sliding, the weight of block C must be equal to or greater than the static friction force, which is 9.8 N.
Therefore, the minimum weight of block C to keep block A from sliding is 9.8 N.