Pleaseee help explain how to draw the free body diagrams for this problem!!!
Two blocks with masses m1 = 10.9 kg and m2 = 79.3 kg, shown in the figure (in the figure the 79.3kg block(m2) is on a flat surface, the 10.9kg(m1) block is pinned against the block with the greater mass and elevated off the surface that m2 is resting on), are free to move. The coefficent of static friction between the blocks is 0.66 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2? in the diagram the force in acting horizontally to the right on m1.
i don't need the whole problem worked out. i just want to understand how the forces acting on the two blocks should be drawn in the free body diagram.
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To draw the free body diagrams for the two blocks, consider the forces acting on each block individually.
For block m1 (10.9 kg), there are three forces to consider:
1. The force of gravity (weight): This force acts downward and has a magnitude of m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2). Draw this force as an arrow pointing downwards from the center of m1.
2. The normal force: This force is the reaction force exerted by m2 on m1. It acts perpendicular to the surface of contact between the two blocks. Since m1 is resting on m2, the normal force acts upwards and has the same magnitude as the weight of m1. Draw this force as an arrow pointing upwards from the center of m1.
3. The static friction force: This force opposes the tendency of m1 to slide horizontally with respect to m2. Since m1 is pinned against m2, the static friction force acts horizontally towards the left, opposing the applied force F. The magnitude of the static friction force can be calculated as the product of the coefficient of static friction (μ) and the normal force. In this case, the magnitude of the static friction force is μ * (m1 * g). Draw this force as an arrow pointing towards the left side of m1.
For block m2 (79.3 kg), there are two forces to consider:
1. The force of gravity (weight): This force acts downward and has a magnitude of m2 * g. Draw this force as an arrow pointing downwards from the center of m2.
2. The normal force: Since the surface beneath m2 is frictionless, the normal force acts vertically upwards and has the same magnitude as the weight of m2. Draw this force as an arrow pointing upwards from the center of m2.