A mass of 11 kg is placed on a frictionless incline which is inclined at an angle of 69 degrees above the horizontal. It is held in place by a rope which is attached to a wall at the top of the incline. The rope is angled 13 degrees above the direction parallel to the incline, however. What is the normal force on the mass in Newtons?
help me solve please!!!!
First draw a diagram of the system (right), then draw the free body diagram (left), see:
http://imageshack.us/photo/my-images/192/1317926143.jpg/
With the forces isolated, weight (mg), tension on the rope (T), and the normal reaction (N), and the angles (in degrees) known, they can be readily resolved into the x and y-components and N calculated.
The equations to resolve forces are:
Vertical (solve for T)
mg=Tcos(8)
Horizontal (solve for N)
Ncos(21)=Tsin(8)
To solve this problem, we need to analyze the forces acting on the mass.
Let's break down the forces acting on the mass on the incline:
1. Weight (W): This is the force due to gravity acting vertically downward. It can be calculated using the formula W = m * g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. Normal Force (N): This force acts perpendicular to the incline and counteracts the component of the weight that is perpendicular to the incline. It can be calculated using the formula N = m * g * cos(angle of incline).
3. Tension Force (T): This is the force provided by the rope, holding the mass in place. It can be resolved into two components: one parallel to the incline and one perpendicular to the incline.
The force diagram looks like this:
|--- N
|
|--- T(parallel to incline)
|
|--- T(perpendicular to incline)
|
v
s
Now, let's calculate the normal force.
Given:
- Mass (m) = 11 kg
- Angle of incline (α) = 69 degrees
- Angle between tension and parallel direction (β) = 13 degrees
- Acceleration due to gravity (g) = 9.8 m/s^2
First, calculate the weight of the object:
W = m * g
W = 11 kg * 9.8 m/s^2
W = 107.8 N
Next, calculate the angle of the tension force with respect to the incline:
θ = α + β
θ = 69 degrees + 13 degrees
θ = 82 degrees
Then, calculate the perpendicular component of the tension force:
T(perpendicular to incline) = T * cos(θ)
T(perpendicular to incline) = W
T(perpendicular to incline) = 107.8 N
Finally, calculate the normal force acting on the mass:
N = m * g * cos(α)
N = 11 kg * 9.8 m/s^2 * cos(69 degrees)
N ≈ 44.6 N
Therefore, the normal force on the mass is approximately 44.6 Newtons.