A mass of 17 kg is placed on a frictionless incline which is inclined at an angle of 59 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 17 degrees above the direction parallel to the incline, however. What is the normal force on the mass in Newtons?
Hmm, based off of what I think.
The rope will be holding the mass at an angle of 76 degrees with respect to a horizontal x- axis.
Fg = 9.81 * 17
= 167N
167N sin 76 <--- thinking about the gravity force being weaker since its at an angle.
=162N
Now calculate the FN
162N cos 59
= 83.4N is your Normal force on the incline.
Sounds right to me, check with your instructor.
To find the normal force on the mass in Newtons, we need to consider the forces acting on the mass.
1. Weight force (mg): The weight force always acts vertically downward and has a magnitude equal to the mass (m) multiplied by the acceleration due to gravity (g = 9.8 m/s^2).
2. Tension force (T): The tension force in the rope keeps the mass in place and acts along the length of the rope. Its magnitude can be resolved into two components: one parallel to the incline and one perpendicular to the incline.
3. Normal force (N): The normal force acts perpendicular to the surface of the incline and prevents the mass from sinking into the incline. This is the force we need to find.
The normal force is equal in magnitude and opposite in direction to the perpendicular component of the tension force. Thus, we need to find the perpendicular component of the tension force.
To find the perpendicular component of the tension force, we need to break it down into its components:
Perpendicular component of tension force (T⊥) = T * sin(θ)
where θ is the angle between the rope and the direction parallel to the incline (17 degrees in this case).
Now that we have the perpendicular component of the tension force, we can set it equal to the normal force:
N = T⊥
To find the value of T, we can decompose it into its components.
Parallel component of tension force (T∥) = T * cos(θ)
Since the incline is frictionless, the parallel component of the tension force is equal in magnitude and opposite in direction to the weight force:
T∥ = mg
Now, we can determine the magnitude of the tension force T:
T = T∥ / cos(θ) = mg / cos(θ)
Substituting this equation back into the equation for the perpendicular component of the tension force, we have:
T⊥ = (mg / cos(θ)) * sin(θ)
Finally, substituting the value of T⊥ into the equation for the normal force, we get:
N = T⊥ = (mg / cos(θ)) * sin(θ)
In this case, m = 17 kg, θ = 59 degrees, and θ = 17 degrees. Plug in these values into the equation to calculate the normal force (N) in Newtons.