a force of magnitude F = 12.0 Napplied to an 8.00 kg block at a downward angle ofθ = 30.0°. The coefficient of static friction between blockand floor is 𝜇s = 0.700; the coefficient of kinetic friction is𝜇k = 0.400. Does the block begin to slide or does it remainstationary? What is the magnitude of the frictional forceon the block?
Question
a force of magnitude F = 12.0 Napplied to an 8.00 kg block at a downward angle ofθ = 30.0°. The coefficient of static friction between blockand floor is 𝜇s = 0.700; the coefficient of kinetic friction is𝜇k = 0.400. Does the block begin to slide or does it remainstationary? What is the magnitude of the frictional forceon the block?
Solution
First, we need to find the components of the force F. The horizontal component (F_x) is Fcos(θ) and the vertical component (F_y) is Fsin(θ).
Given: F = 12.0 N θ = 30.0° m = 8.00 kg g = 9.8 m/s² (acceleration due to gravity) μs = 0.700 (coefficient of static friction) μk = 0.400 (coefficient of kinetic friction)
Step 1: Calculate the components of the force F F_x = Fcos(θ) = 12.0 N * cos(30.0°) = 10.39 N F_y = Fsin(θ) = 12.0 N * sin(30.0°) = 6.00 N
Step 2: Calculate the normal force (N). The normal force is the sum of the weight of the block (mg) and the vertical component of the force (F_y). N = mg + F_y = 8.00 kg * 9.8 m/s² + 6.00 N = 84.4 N
Step 3: Calculate the maximum static friction force (f_s_max) using the formula f_s_max = μs * N f_s_max = 0.700 * 84.4 N = 59.08 N
Step 4: Compare the horizontal component of the force (F_x) with the maximum static friction force (f_s_max). If F_x > f_s_max, the block will start to slide. If F_x ≤ f_s_max, the block will remain stationary. Since F_x (10.39 N) is less than f_s_max (59.08 N), the block will remain stationary.
Step 5: Since the block is not moving, the frictional force is equal to the horizontal component of the force (F_x), which is 10.39 N.
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