While two forces act on it, a particle is to move at the constant velocity v = (3 m/s) i - (4 m/s) j. One of the forces is F1 = (2 N) i + (- 6 N) j. What is the other force?
( _____ N) i + ( _____ N) j
Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 6.9 N, acting 70° north of west. What is the magnitude of the body's acceleration?
on the first, constant velocity means no net force. So the other force will be the negative of the first.
Second: add the forces
resultantmagnitude=sqrt (9^2+.9^2)
To determine the other force acting on the particle with a known velocity, we can use Newton's second law of motion which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, since the particle is moving at a constant velocity, we know that there is no net force acting on it. Therefore, the sum of the forces acting on the particle must be equal to zero.
Let's label the unknown force F2 = (x N) i + (y N) j.
Since the velocity is given as v = (3 m/s)i - (4 m/s)j and the known force F1 = (2 N)i + (-6 N)j, we can set up the equation:
F1 + F2 = 0
(2 N)i + (-6 N)j + (x N)i + (y N)j = 0
This equation can be rewritten as:
(x + 2)i + (y - 6)j = 0
By comparing the coefficients of the i and j terms on both sides of the equation, we get two separate equations:
x + 2 = 0 (1)
y - 6 = 0 (2)
Solving equations (1) and (2), we find:
x = -2
y = 6
Therefore, the other force acting on the particle is:
(-2 N) i + (6 N) j
Now let's move on to the second question:
To find the magnitude of the body's acceleration, we need to first determine the net force acting on the body.
Let's resolve the given forces into their horizontal (x) and vertical (y) components:
Horizontal force due east = 9.0 N
Vertical force 70° north of west = 6.9 N
Using trigonometry, we can calculate the x and y components of the vertical force:
x-component = 6.9 N * cos(70°) = 6.9 N * 0.342 = 2.364 N (negative because it's directed west)
y-component = 6.9 N * sin(70°) = 6.9 N * 0.940 = 6.486 N (positive because it's directed north)
Now, let's determine the net x and y components of the forces:
Net x-component = 9.0 N + 2.364 N = 11.364 N (directed east)
Net y-component = 6.486 N (no other vertical forces are acting)
Since the body can move over a frictionless floor, the horizontal forces do not affect its acceleration. The acceleration will only be due to the net vertical force.
Using Newton's second law in the y-direction:
Net force (Fnet) = mass (m) * acceleration (a)
Fnet = Net y-component of forces = 6.486 N
m = mass of the body = 3.0 kg
Fnet = m * a
6.486 N = 3.0 kg * a
Solving for a:
a = 6.486 N / 3.0 kg
a = 2.162 m/s^2
Therefore, the magnitude of the body's acceleration is 2.162 m/s^2.