Answer:
See explanation for step by step procedure to get answer.
Explanation:
Given that:
The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.
See the attachments for complete steps to get answer.
At equilibrium, the sum of clockwise and anticlockwise moments about a point is zero
The mass of that balances the system is kg
The normal reaction force at the fulcrum is <u>5,804.25 N</u>
Rason:
The mass of the stick = 150.0 g
Mass m₁ on the left = 50.0 g, location = 30 cm to the left of m₂
Mass m₂ on the left = 75.0 g, location = 40 cm to the left of the fulcrum
Mass m₃ on the right of the fulcrum. location = 30 cm to the right of the fulcrum
Required:
To find the mass of m₃
Solution:
Taking moment about the fulcrum, we have;
50 × (30 + 40) + 75 × (40) + 150 × 20 = m₃ × 30
9,500 = m₃×30
The mass of that balances the system when it is attached at the right end of the stick, m₃ = kg
Normal reaction at the fulcrum = (50 + 75 + 150 + ) × 9.81 = 5804.25
The normal reaction at the fulcrum is 5,804.25 N
Learn more about the moment of a force here:
