A 4.00-kg block rests on a 30.0° incline. If the coefficient of static friction between the block and the incline is 0.700, what magnitude horizontal force F must act on the block to start it moving up the incline, given g= 9.80m/s^2?
To find the magnitude of the horizontal force F required to start the block moving up the incline, we need to consider the forces acting on the block.
1. Start by drawing a free-body diagram of the block on the incline. Label the following forces:
- The gravitational force acting vertically downwards, which can be divided into two components: mg * cosθ (the component parallel to the incline) and mg * sinθ (the component perpendicular to the incline).
- The normal force (N) exerted by the incline on the block perpendicular to the incline.
- The static friction force (fs) acting parallel to the incline and opposing the motion of the block.
2. Now, solve for the component of the gravitational force parallel to the incline, which is mg * sinθ.
- mg = mass * acceleration due to gravity = 4.00 kg * 9.80 m/s^2 = 39.2 N
- sinθ = sin(30.0°) = 0.5
Therefore, the parallel component of the gravitational force is 39.2 N * 0.5 = 19.6 N.
3. Determine the maximum static friction force (fsmax) that can be applied to the block without causing it to move.
- fsmax = coefficient of static friction * normal force
- The normal force can be determined using N = mg * cosθ.
cosθ = cos(30.0°) = 0.866
N = 4.00 kg * 9.80 m/s^2 * 0.866 = 33.92 N
fsmax = 0.700 * 33.92 N = 23.744 N
4. The horizontal force F required to start the block moving up the incline should overcome the maximum static friction force (fsmax) acting on the block.
- F ≥ fsmax
Therefore, F ≥ 23.744 N.
Hence, the magnitude of the horizontal force F required to start the block moving up the incline is at least 23.744 N.