Determine the horizontal force P required to start the movement of the 200 lb blockup the 300inclined surface. The static coefficient of friction between the inclined surface and the block is ms = 0.25.
To determine the horizontal force P required to start the movement of the 200 lb block up the 30° inclined surface, we need to consider the forces acting on the block.
First, we need to find the weight of the block. The weight is the mass of the block multiplied by the acceleration due to gravity. Given that the mass of the block is 200 lb, we can convert it to slugs by dividing by the gravitational constant (32.2 ft/s^2):
Weight of the block = mass × acceleration due to gravity
= 200 lb / 32.2 ft/s^2
≈ 6.21 slugs (rounded to two decimal places)
Now let's consider the forces acting on the block on the inclined surface.
1. Weight (W): This force is acting vertically downward and can be calculated as:
W = weight of the block × g
= 6.21 slugs × 32.2 ft/s^2
≈ 200 lb (rounded to the nearest pound)
2. Normal Force (N): This force is perpendicular to the inclined surface. It can be calculated as:
N = weight of the block × cos(angle of the incline)
= 200 lb × cos(30°)
≈ 173 lb (rounded to the nearest pound)
3. Friction Force (f): This force opposes the motion of the block and acts parallel to the inclined surface. The static friction force can be calculated as:
f = static coefficient of friction (ms) × N
= 0.25 × 173 lb
≈ 43 lb (rounded to the nearest pound)
Now, using the free-body diagram, we can analyze the forces acting on the block along the incline:
/|\
/ | \ P
/ | \
/ | \
--------------
N f W
To start the movement of the block up the inclined surface, the horizontal force P needs to overcome the static friction force f. Therefore, the magnitude of the horizontal force P is equal to the friction force f:
P = f
= 43 lb (rounded to the nearest pound)
So, the horizontal force P required to start the movement of the 200 lb block up the 30° inclined surface is approximately 43 lb.