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-8. A block that weighs 60 N is initially at rest on a horizontal surface. A horizontal force F, of magnitude 22 N and a vertical force F, are then applied to the block. The coefficients of friction for the block and the surface are u, = 0.50 and µk 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of F2 is (a) 10 N, (b) 20 N F2

-8. A block that weighs 60 N is initially at rest on a horizontal surface. A horizontal force F, of magnitude 22 N and a vertical force F, are then applied to the block. The coefficients of friction for the block and the surface are u, = 0.50 and µk 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of F2 is (a) 10 N, (b) 20 N F2

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### Problem 8: A block that weighs 60 N is initially at rest on a horizontal surface. A horizontal force \( F_1 \) of magnitude 22 N and a vertical force \( F_2 \) are then applied to the block. The coefficients of friction for the block and the surface are \( \mu_s = 0.50 \) and \( \mu_k = 0.25 \). Determine the magnitude of the frictional force acting on the block if the magnitude of \( F_2 \) is: (a) 10 N (b) 20 N #### Diagram: The diagram shows a block resting on a horizontal surface. Two forces are acting on the block: - \( \mathbf{F_1} \) is a horizontal force acting to the right. - \( \mathbf{F_2} \) is a vertical force acting upwards from the center of the block. #### Explanation of Forces: 1. **Block Weight (W):** The force due to gravity on the block, which is 60 N, acts vertically downward. 2. **Applied Horizontal Force (\( F_1 \)):** A force of 22 N acts horizontally to the right. 3. **Applied Vertical Force (\( F_2 \)):** - Case (a): \( F_2 = 10 \) N - Case (b): \( F_2 = 20 \) N 4. **Frictional Force:** The force opposing the motion of the block due to contact with the surface. The frictional force can be calculated using the normal reaction and the coefficients of friction. #### Coefficients of Friction: - Static Friction Coefficient (\( \mu_s \)): 0.50 - Kinetic Friction Coefficient (\( \mu_k \)): 0.25
Transcribed Image Text:### Problem 8: A block that weighs 60 N is initially at rest on a horizontal surface. A horizontal force \( F_1 \) of magnitude 22 N and a vertical force \( F_2 \) are then applied to the block. The coefficients of friction for the block and the surface are \( \mu_s = 0.50 \) and \( \mu_k = 0.25 \). Determine the magnitude of the frictional force acting on the block if the magnitude of \( F_2 \) is: (a) 10 N (b) 20 N #### Diagram: The diagram shows a block resting on a horizontal surface. Two forces are acting on the block: - \( \mathbf{F_1} \) is a horizontal force acting to the right. - \( \mathbf{F_2} \) is a vertical force acting upwards from the center of the block. #### Explanation of Forces: 1. **Block Weight (W):** The force due to gravity on the block, which is 60 N, acts vertically downward. 2. **Applied Horizontal Force (\( F_1 \)):** A force of 22 N acts horizontally to the right. 3. **Applied Vertical Force (\( F_2 \)):** - Case (a): \( F_2 = 10 \) N - Case (b): \( F_2 = 20 \) N 4. **Frictional Force:** The force opposing the motion of the block due to contact with the surface. The frictional force can be calculated using the normal reaction and the coefficients of friction. #### Coefficients of Friction: - Static Friction Coefficient (\( \mu_s \)): 0.50 - Kinetic Friction Coefficient (\( \mu_k \)): 0.25
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