The diagram shows a block of mass m = 3.50 kg resting on a plane inclined at an angle of θ = 40° to the horizontal. The coefficient of static friction between the block and the plane is μstatic = 0.30, and the block is stationary but just on the point of sliding down the slope.
Diagram
The diagram shows the four forces acting on the block: an applied force F1 acting perpendicular to the slope (in the –y direction), the block's weight mg, the normal reaction force N and the force of static friction, Ff. In this case, the force of static friction acts up the slope, opposing the tendency of the block to move down the slope.
Find the the minimum magnitude of the applied force F1 that can be exerted if the block is to remain stationary.
M*g = 3.5 * 9.8 = 34.3 N. = Wt. of block.
34.3*sln40 = 22 N. = Force parallel to plane.
34.3*Cos40 = 26.3 N. = Normal force.
Ff = 0.3 * 26.3 = 7.88 N. = Force of static friction.
F1 - Ff = M*a.
F1 - 7.88 = 3.5*0,
F1 = 7.88 N. = Force applied.
Correction: F1 - 7.88 - 22 = 3.5*0.
F1 = 29.88 N.
To find the minimum magnitude of the applied force F1 that can be exerted if the block is to remain stationary, we need to consider the forces acting on the block.
1. Identify the forces:
- The weight of the block, mg, acting vertically downwards.
- The normal reaction force, N, acting perpendicular to the plane.
- The force of static friction, Ff, acting up the slope.
- The applied force, F1, acting perpendicular to the slope (in the –y direction).
2. Resolve the weight force:
Since the plane is inclined at an angle of θ = 40° to the horizontal, we can resolve the weight force into two components: the force acting down the slope, mg * sin(θ), and the force acting perpendicular to the slope (normal force), mg * cos(θ).
3. Determine the normal force:
The normal force, N, is equal to the component of the weight force that is perpendicular to the slope, which is mg * cos(θ).
4. Calculate the force of static friction:
The force of static friction, Ff, opposes the tendency of the block to slide down the slope. Its magnitude is given by the equation Ff ≤ μstatic * N, where μstatic is the coefficient of static friction.
In this case, the block is stationary but just on the point of sliding down the slope, so Ff = μstatic * N.
5. Equilibrium condition:
For the block to remain stationary, the forces in the vertical direction (y-direction) must balance each other. This means that the sum of the vertical forces must be zero. Therefore, we have:
N - mg * sin(θ) = 0 (Since the block is not moving vertically)
6. Equilibrium condition in the horizontal direction (x-direction):
For the block to remain stationary, the forces in the horizontal direction (x-direction) must balance each other. This means that the sum of the horizontal forces must be zero. Therefore, we have:
F1 - Ff - mg * cos(θ) = 0 (Since the block is not moving horizontally)
7. Substitute the expression for Ff:
From step 4, we have Ff = μstatic * N. Substituting this into the equilibrium equation from step 6, we get:
F1 - μstatic * N - mg * cos(θ) = 0
8. Substitute the expression for N:
From step 3, we have N = mg * cos(θ). Substituting this into the equilibrium equation from step 7, we get:
F1 - μstatic * mg * cos(θ) - mg * cos(θ) = 0
9. Simplify and solve for F1:
We can now solve the equation for F1:
F1 = μstatic * mg * cos(θ) + mg * cos(θ)
F1 = (μstatic + 1) * mg * cos(θ)
Substituting the given values: m = 3.50 kg, θ = 40°, and μstatic = 0.30, we can calculate the minimum magnitude of the applied force F1 that can be exerted.