1)An object is suspended from two points on the ceiling 25cm apart by two pieces of rope of lengths 27cm and 24 if the object exerts from a downward force 200N then find the tension in each rope
2)An airplane heading north at 550km/h encounters a wind of 175km/h from 32 degrees of north of east,determine the resultant ground velocity of the plane and the direction of the plane
3)A sailor climbs a mast at 0.5m/s on a ship travelling north at 12m/s while the current flows east at 3m/s what is the speed of sailor relative to ocean floor
1) To find the tension in each rope, we can use the concept of equilibrium. In this case, the object is suspended from two points, and the tension in each rope acts in a different direction.
Let's assume that one rope is attached to the left point and the other rope is attached to the right point. We'll call the tension in the left rope T1 and the tension in the right rope T2.
Using the concept of equilibrium, we know that the sum of all forces acting on the object must be zero. In this case, the object exerts a downward force of 200N, so we also have an upward force of 200N acting on the object.
Now, let's analyze the forces acting in the horizontal and vertical directions separately:
In the vertical direction:
The total upward force from the tensions in the ropes (T1 and T2) must balance out the downward force of 200N. Therefore, T1 + T2 = 200N.
In the horizontal direction:
Since the object is stationary, there is no net force acting in the horizontal direction. Therefore, the horizontal components of the tensions in the ropes must be equal. Let's call the angle between the horizontal direction and the left rope angle θ1, and the angle between the horizontal direction and the right rope θ2. We can use trigonometry to relate the horizontal components of the tension to the angles:
T1 * cos(θ1) = T2 * cos(θ2).
Now we have two equations:
T1 + T2 = 200N (from the vertical equilibrium)
T1 * cos(θ1) = T2 * cos(θ2) (from the horizontal equilibrium)
We can now solve these two equations simultaneously to find the tension in each rope.
2) To determine the resultant ground velocity of the plane and the direction, we need to consider the vector addition of the airplane's velocity and the wind's velocity.
The airplane's velocity, heading north at 550 km/h, can be represented as a vector pointing straight up. This vector has a magnitude of 550 km/h and an angle of 90 degrees (measured counterclockwise from the positive x-axis).
The wind's velocity, with a speed of 175 km/h and coming from 32 degrees east of north, can be represented as a vector pointing in the direction of 32 degrees counterclockwise from the positive y-axis. This vector has a magnitude of 175 km/h and an angle of (90 + 32) degrees.
To find the resultant ground velocity, we add these two vectors together. We can do this by splitting each vector into its horizontal and vertical components.
For the airplane's velocity, the horizontal component is 0 km/h (since it is moving purely north) and the vertical component is 550 km/h.
For the wind's velocity, the horizontal component can be found using trigonometry: 175 km/h * sin(32) = 93.622 km/h. The vertical component can be found using trigonometry as well: 175 km/h * cos(32) = 149.956 km/h.
Now, add the horizontal components and vertical components separately to find the resultant ground velocity:
Horizontal component: 0 km/h + 93.622 km/h = 93.622 km/h.
Vertical component: 550 km/h + 149.956 km/h = 699.956 km/h (this will be the magnitude of the resultant velocity).
The direction of the resultant velocity can be found using trigonometry. The angle with respect to the positive x-axis, θ, can be found using the inverse tangent function:
θ = tan^(-1)(Vertical component / Horizontal component)
θ = tan^(-1)(699.956 km/h / 93.622 km/h)
θ ≈ 76.6 degrees.
Therefore, the resultant ground velocity of the plane is approximately 699.956 km/h at an angle of 76.6 degrees counterclockwise from the positive x-axis.
3) To find the speed of the sailor relative to the ocean floor, we need to consider the vector addition of the sailor's velocity and the ship's velocity relative to the ocean floor.
The sailor is climbing the mast at a speed of 0.5 m/s in a north direction. The ship is traveling north at a speed of 12 m/s. Additionally, there is a current flowing east at a speed of 3 m/s.
To find the relative speed of the sailor to the ocean floor, we can treat the ship's velocity and the current's velocity as vectors and subtract them from the sailor's velocity.
The sailor's velocity is given as 0.5 m/s north. The ship's velocity is given as 12 m/s north, and the current's velocity is given as 3 m/s east.
To subtract these vectors, we can break them down into their horizontal and vertical components.
- The sailor's velocity has no horizontal component (0 m/s) and a vertical component of 0.5 m/s.
- The ship's velocity has no horizontal component (0 m/s) and a vertical component of 12 m/s.
- The current's velocity has a horizontal component of 3 m/s and no vertical component (0 m/s).
Now, subtract the horizontal and vertical components separately:
Horizontal component: 0 m/s - 3 m/s = -3 m/s.
Vertical component: 0.5 m/s - 12 m/s = -11.5 m/s.
The resultant velocity is approximately -3 m/s in the east direction and -11.5 m/s in the downward direction (or south). The magnitude of this resultant velocity can be found using the Pythagorean theorem:
Resultant speed = sqrt((-3 m/s)^2 + (-11.5 m/s)^2)
Resultant speed ≈ 11.9 m/s.
Therefore, the speed of the sailor relative to the ocean floor is approximately 11.9 m/s, and it is directed 14.5 degrees below the horizontal (south direction).