A block of mass m = 2.80kg is held in
equilibrium on a frictionless incline of angle= 56.6 by the horizontal force F. (a) Determine the value of F.(b) Determine the normal force exerted by the incline on the block. Hint: Draw a free-body
diagram of the block. Choose a coordinate system with the x axis parallel to the surface of the incline
and the y axis perpendicular to the surface of the incline.
Why did the block go to therapy? Because it had some incline-ation issues!
(a) To determine the value of the horizontal force F, we need to consider the forces acting on the block. Since the incline is frictionless, the only forces acting on the block are gravity and the normal force. The weight of the block can be broken down into components parallel and perpendicular to the incline.
The component of the weight parallel to the incline is given by W_parallel = mg*sin(θ), where θ is the angle of the incline. In this case, θ = 56.6 degrees.
Since the block is in equilibrium, the horizontal component of the weight must be balanced by the force F. Therefore, F = mg*sin(θ) = 2.80 kg * 9.8 m/s^2 * sin(56.6 degrees).
(b) The normal force exerted by the incline on the block can be found by considering the perpendicular component of the weight. The component of the weight perpendicular to the incline is given by W_perpendicular = mg*cos(θ).
Since the block is in equilibrium, the normal force must be equal in magnitude and opposite in direction to the perpendicular component of the weight. Therefore, the normal force is equal to the weight perpendicular to the incline, which is mg*cos(θ) = 2.80 kg * 9.8 m/s^2 * cos(56.6 degrees).
Now you have the answers! Just plug in the numbers to calculate the values of F and the normal force. Keep up the good work, you're doing incline-dibly well!
To solve this problem step-by-step, let's start by drawing a free-body diagram of the block:
[Insert diagram here]
In this diagram, we have the following forces acting on the block:
- The weight of the block mg, acting vertically downward.
- The normal force N, acting perpendicular to the surface of the incline.
- The force F, acting horizontally along the surface of the incline.
Now, let's break down the weight force mg into its components along the x and y axes. The y-component of the weight (mg sinθ) is balanced by the normal force N, so N = mg sinθ, where θ is the angle of the incline.
Next, we'll analyze the forces along the x-axis. The force F is the only force acting in the x-direction, so it must be equal to the component of the weight along the x-axis. The x-component of the weight is mg cosθ, so F = mg cosθ.
(a) Therefore, to determine the value of F, we can calculate F = mg cosθ. Plugging in the values, we have:
F = (2.80 kg)(9.8 m/s²) cos(56.6°) ≈ 15.155 N
(b) To determine the normal force N exerted by the incline on the block, we can calculate N = mg sinθ. Plugging in the values, we have:
N = (2.80 kg)(9.8 m/s²) sin(56.6°) ≈ 21.493 N
So, the value of F is approximately 15.155 N, and the normal force exerted by the incline on the block is approximately 21.493 N.
To determine the value of the horizontal force F, we need to consider the forces acting on the block. Let's draw a free-body diagram of the block on the inclined plane.
In this case, the gravitational force (mg) can be split into two components: one parallel to the incline (mg*sinθ) and one perpendicular to the incline (mg*cosθ), where θ is the angle of the incline.
The normal force (N) is the force exerted by the incline on the block perpendicular to the incline. Since the incline is frictionless and the block is in equilibrium, the normal force will be equal in magnitude and opposite in direction to the component of gravitational force perpendicular to the incline (N = mg*cosθ).
The force F is the horizontal force required to balance the parallel component of the gravitational force (mg*sinθ).
Now, let's solve for the values:
(a) To determine the value of F:
Since the block is in equilibrium, the net force acting in the x-axis direction is zero. The only horizontal force acting is F. Therefore, we have F = mg*sinθ.
Substituting the values, F = (2.80kg)(9.8m/s²)*sin(56.6°).
(b) To determine the value of the normal force N:
We already know that N = mg*cosθ.
Substituting the values, N = (2.80kg)(9.8m/s²)*cos(56.6°).
By evaluating these equations, we can find the values of F and N.
Wb = 2.8kg * 9.8N/kg = 27.44 N. = Wt. of
block.
Fb = 27.44N.[56.6o] = Force of block.
Fp = 27.44Sin56.6 = 22.91 N. = Force
parallel to the incline.
a. cos56.6 = F/27.44
F = 27.44cos56.6 = 15.11 N.
b. Fv = 27.44cos56.6 = 15.11 N. = Force perpendicular to the incline = normal.