A 10N block is attached to the wall by a rope on a frictionless surface. If the lower surface makes an angle of 30 degrees with the horizontal, calculate the (a) normal force (b) tension on the rope.
To calculate the normal force (N) and the tension on the rope (T), we need to break down the forces acting on the block.
Let's start by drawing a free-body diagram of the block:
⇡
│ N
│
─────╱│╲─────
╱ ┼ ──╲
╱│ ╲
╱ │T ╲
╱ │ ╲
m╱ │ ╲
╲ │ ╲
─────╲│╱─────
╱ ┼ --→ Fg
╱ │
ϴ ╱ │
╱ │
╱ │
╱ │
─→ F
In the diagram, N represents the normal force, T represents the tension in the rope, Fg represents the force due to gravity, and ϴ represents the angle between the horizontal surface and the incline.
We know that the weight (force due to gravity) acting on an object is given by the formula:
Fg = mg
where m is the mass of the object and g is the acceleration due to gravity.
In this case, we are given that the weight (Fg) is 10N.
To find the normal force (N), we can use trigonometry. The normal force acts perpendicular to the inclined surface, so it can be calculated using the formula:
N = Fg · cos(ϴ)
where ϴ is the angle between the vertical direction and the inclined surface.
In this case, the angle is given as 30 degrees. Plugging in the values, we get:
N = 10N · cos(30°)
Calculating the value of cos(30°), we find that it is √3/2. So the normal force is:
N = 10N · √3/2
To find the tension in the rope (T), we can use trigonometry as well. The tension acts along the inclined surface and can be calculated using the formula:
T = Fg · sin(ϴ)
Plugging in the values, we get:
T = 10N · sin(30°)
Calculating the value of sin(30°), we find that it is 1/2. So the tension in the rope is:
T = 10N · 1/2
Simplifying, we get:
(a) Normal force (N) = 10N · √3/2
(b) Tension in the rope (T) = 10N · 1/2
So, the normal force is 5√3 N and the tension in the rope is 5 N.