A block sits against a vertical wall and a 120 lb force P acts diagonally to the left on the block at an angle below the horizontal. Determine the magnitude and direction of the friction force which the vertical wall exerts on the 100 lb block if theta is (a)15 degrees (b)30 degrees
For part a I got that the answer is 46.36 lb up but I can't get part b to work correctly.
You have provided incomplete information.
Were you given a value for the coefficient of friction? From what direction is theta measured?
yeah the coefficient of static friction is 0.5 and coeffient of kinetic friction is 0.4. theta is measured below the horizontal
To determine the magnitude and direction of the friction force exerted by the vertical wall on the block, you can break down the forces acting on the block and use Newton's second law (F = ma) to solve for the friction force.
Let's analyze the situation step by step and work through both parts of the problem:
Part a) When theta is 15 degrees:
1. Draw a diagram of the forces acting on the block:
- There is a vertical force acting downward due to the weight of the block (100 lb).
- The 120 lb force P acts diagonally to the left at an angle below the horizontal.
- The friction force, Ff, is exerted by the wall on the block and opposes the motion sideways.
2. Resolve the weight and force P into their vertical and horizontal components:
- The vertical component of the weight is W = 100 lb, acting downward.
- The horizontal component of force P is Px = P * cos(theta).
3. Calculate the net horizontal force on the block:
- Fnet = Px - Ff, where Fnet is the net force acting horizontally.
- Since the block is not accelerating horizontally (in equilibrium), Fnet is zero.
- Therefore, Px = Ff.
4. Calculate the friction force, Ff:
- Ff = Px = P * cos(theta) = 120 lb * cos(15 degrees) = 120 lb * 0.9659 ≈ 115.91 lb.
5. Determine the direction of the friction force:
- Since the friction force opposes the horizontal motion, it acts to the right (opposite to the direction of P).
Hence, in part a, the magnitude of the friction force is approximately 115.91 lb, and it acts to the right.
Now, let's move on to part b) when theta is 30 degrees:
1. Repeat steps 1 and 2 from part a to draw the diagram and resolve the weight and force P into their components.
- Vertical component of the weight: W = 100 lb, downward.
- Horizontal component of force P: Px = P * cos(theta).
2. Calculate the net horizontal force on the block:
- Fnet = Px - Ff.
- Since the block is not accelerating horizontally (in equilibrium), Fnet is zero.
- Therefore, Px = Ff.
3. Calculate the friction force, Ff:
- Ff = Px = P * cos(theta) = 120 lb * cos(30 degrees) = 120 lb * 0.866 ≈ 103.92 lb.
4. Determine the direction of the friction force:
- Since the friction force opposes the horizontal motion, it acts to the right (opposite to the direction of P).
Hence, in part b, the magnitude of the friction force is approximately 103.92 lb, and it acts to the right.
Therefore, the answer for Part b is a magnitude of 103.92 lb to the right.