Block A w/ Friction: Find Force & Max Weight

[amm617]

Homework Statement

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]

Homework Equations Ff=μNThe Attempt at a Solution a)Ff=μN Ff=.30(64.2N) Ff=19.26N b)The maximum weight for which the system will remain in equilibrium is the magnitude of the normal force (N) exerted on block A. The normal force is equal to the weight of the block (64.2N) minus the downward force of the 11.3N weight (i.e., 64.2-11.3=52.9N). Therefore, the maximum weight for which the system will remain in equilibrium is 52.9N.

 

[Moetasim]

a) The friction force exerted on block A can be calculated using the formula Ff = μN, where μ is the coefficient of static friction and N is the normal force exerted by the surface on the block. In this case, the normal force is equal to the weight of the block, which is 64.2N. Therefore, the friction force is Ff = 0.30 x 64.2N = 19.26N.

b) To find the maximum weight for which the system will remain in equilibrium, we need to consider the forces acting on the block in the vertical direction. These forces are the weight of the block (64.2N), the weight attached to the string (11.3N), and the vertical component of the tension in the string (Tsinθ, where θ is the angle of the string with respect to the horizontal). Since the system is in equilibrium, the sum of these forces must be equal to zero.

Therefore, we can write the equation: 64.2N + 11.3N + Tsinθ = 0

Solving for T, we get T = -75.5N. Since the tension in the string cannot be negative, the maximum weight that can be supported by the system is 75.5N. This means that if the weight on the string exceeds 75.5N, the system will no longer be in equilibrium and the block will start to move.