How Much Weight Can Block A Support Before Sliding?

[amm617]

Block A in the figure weighs 64.2N . The coefficient of static friction between the block and the surface on which it rests is 0.30. The weight is 11.3N and the system is in equilibrium.

The picture is a block A, sitting on a table. A string is attached horizontally and then the string breaks off into two separate strings. one connects to the wall with a angle of 45 degrees upward. and the other goes straight down with a weight on the end of it that weighs 11.3 N.

a)Find the friction force exerted on block A.
b)Find the maximum weight for which the system will remain in equilibrium.

 

[Nate810]

a) The friction force exerted on block A is 19.26N. This can be calculated using the formula Ff=μ*Fn, where μ is the coefficient of static friction and Fn is the normal force. The normal force can be found by subtracting the weight of the system (11.3N) from the weight of the block (64.2N). Therefore, the normal force is 52.9N and the friction force is 19.26N (52.9N x 0.30 = 19.26N). b) The maximum weight for which the system will remain in equilibrium is 86.5N. This can be calculated by taking the sum of the weight of the block (64.2N), the weight of the system (11.3N), and the maximum friction force (19.26N). Therefore, the maximum weight is 86.5N (64.2N + 11.3N + 19.26N = 86.5N).

 

[Moetasim]

a) According to Newton's second law, the sum of all forces acting on an object must be equal to the mass of the object multiplied by its acceleration. In this case, the block is in equilibrium, so the sum of all forces must be equal to zero. The only forces acting on the block are its weight (64.2N) and the tension in the string connected to the wall (T). Therefore, we can set up the following equation:

ΣF = 0
-T + 64.2N = 0
T = 64.2N

Since the tension in the string is equal to the friction force, we can conclude that the friction force exerted on block A is 64.2N.

b) To find the maximum weight for which the system will remain in equilibrium, we need to consider the maximum possible value for the tension in the string. This occurs when the block is on the verge of sliding, which happens when the force of static friction is at its maximum value. The maximum force of static friction can be calculated by multiplying the coefficient of static friction (0.30) by the normal force (64.2N), which gives us 19.26N.

Therefore, the maximum weight that can be added to the system without causing the block to slide is 19.26N. Any weight greater than this will cause the block to slide and the system to no longer be in equilibrium.