A 20 kg wagon is pulled along the level ground by a rope inclined at 30 deg above horizontal. A frictional force of 30 N opposes the motion, how large is the pulling force if the wagon is accelerating at the rate of 0.40m/s2?
F cos 30 - 30 = m a = 20 * 0.40
so
F = [ 30 + 8 ] / 0.866
To find the pulling force, we need to consider the forces acting on the wagon.
First, let's draw a free-body diagram to visualize the forces involved:
-------------------------
| \
____|____ \ <-- Rope
/ \ \
/ Wagon \ \
/ \ \
/ \ \
<---- 30° \ 30 N <--- Friction force
\ /
\ /
\_______/
In the diagram:
- Wagon represents the 20 kg mass of the object being pulled.
- The upward angled arrow represents the pulling force.
- The downward angled arrow represents the frictional force.
- The straight horizontal arrow represents the horizontal component of the rope tension.
From the given information, we know that:
- Mass (m) of the wagon = 20 kg
- Frictional force (F_friction) = 30 N
- Acceleration (a) = 0.40 m/s^2
- Angle (θ) between the inclined rope and the horizontal ground = 30°
Now, let's break down the forces along the horizontal and vertical directions:
1. Vertical direction (upward is positive):
- Tension in the rope = T (unknown)
- Vertical component of tension = T * sin(30°) = T_vertical
Since there is no vertical acceleration (the wagon is on level ground), the vertical forces are balanced:
Sum of vertical forces = 0
Therefore:
T_vertical = Weight of the wagon (m * g) (opposite direction of the upward arrow)
T_vertical = 20 kg * 9.8 m/s^2 = 196 N
2. Horizontal direction (forward is positive):
- Tension in the rope = T (unknown)
- Horizontal component of tension = T * cos(30°) = T_horizontal
- Frictional force = F_friction = 30 N (opposite direction)
We can set up the equation for horizontal forces:
Sum of horizontal forces = ma (from Newton's second law)
T_horizontal - F_friction = m * a
T_horizontal - 30 N = 20 kg * 0.40 m/s^2
T_horizontal - 30 N = 8 N
Now, we know the horizontal component of tension:
T_horizontal = 8 N + 30 N
T_horizontal = 38 N
Finally, to find the pulling force, we need to consider both horizontal and vertical components:
Pulling force = Square root( T_horizontal^2 + T_vertical^2 )
Pulling force = Square root( (38 N)^2 + (196 N)^2 )
Pulling force = Square root( 1444 N^2 + 38416 N^2 )
Pulling force = Square root( 39860 N^2 )
Pulling force ≈ 199.6 N
Therefore, the pulling force required is approximately 199.6 Newtons.