A horizontal force presses the block of mass 10kg to the wall. What is the minimum magnitude of the applied force which can hold the block on the wall without accelerating it down if the coefficient of static friction (pico per second) between the block and the wall is 0.40.
force of gravity down = m g = 98 N
friction force up = mu F = .4 F
so
.4 F = 98
F = 98/.4
To find the minimum magnitude of the applied force that can hold the block on the wall without accelerating it down, we need to consider the forces acting on the block and use the concept of static friction.
Let's break down the forces acting on the block:
1. Weight (mg): This is the force exerted by gravity on the block and acts vertically downward. The weight can be calculated using the formula: weight = mass × acceleration due to gravity (g). In this case, as the mass is 10 kg and the acceleration due to gravity is approximately 9.8 m/s², weight = 10 kg × 9.8 m/s² = 98 N.
2. Normal force (N): This force is exerted by the wall on the block and acts perpendicular to the wall's surface. Since the block is pressed against the wall and not moving vertically, the normal force is equal in magnitude to the weight of the block, i.e., N = 98 N.
3. Applied force (F): This is the horizontal force being applied to the block. We need to find the minimum magnitude of this force to hold the block on the wall without accelerating it down.
4. Static friction force (f_s): This force is exerted by the wall on the block parallel to the wall's surface, opposing the impending motion. The maximum value of static friction can be calculated using the formula: f_s = coefficient of static friction × N. In this case, the coefficient of static friction is given as 0.40, so f_s = 0.40 × 98 N = 39.2 N.
To hold the block on the wall without accelerating it down, the applied force should exactly balance the maximum value of static friction. Therefore, the minimum magnitude of the applied force is equal to the maximum static friction force, which is 39.2 N.