A crate, which has a mass of m= 45 kg, is being accelerated at 3.2m/s^-2 up an frictionless inclined plane, which meets the horizontal at an angle of 35 degrees, by a rope.

What is the magnitude of the normal force acting on the crate?
What is the tension acting in the rope?

N=mgcosα

ma=T-mgsinα =>
T=m(a+gsinα)

To find the magnitude of the normal force acting on the crate, we need to consider the forces acting on the crate in the vertical direction.

First, we need to resolve the weight of the crate into its components parallel and perpendicular to the inclined plane.

1. Weight of the crate (W) = mass (m) x acceleration due to gravity (g)
W = 45 kg x 9.8 m/s^2
W = 441 N

2. The component of the weight parallel to the inclined plane (W_parallel) = W x sin(θ)
W_parallel = 441 N x sin(35°)
W_parallel = 251.46 N

Since the crate is being accelerated upwards, the magnitude of the normal force (N) will be less than the weight of the crate.

3. Magnitude of the normal force (N) = weight of the crate (W) - component of the weight parallel to the inclined plane (W_parallel)
N = W - W_parallel
N = 441 N - 251.46 N
N = 189.54 N

Therefore, the magnitude of the normal force acting on the crate is 189.54 N.

To find the tension acting in the rope, we need to consider the forces acting on the crate in the horizontal direction.

The net force in the horizontal direction is equal to the mass of the crate times the acceleration.

4. Net Force (F_net) = mass (m) x acceleration (a)
F_net = 45 kg x 3.2 m/s^2
F_net = 144 N

Since there is no friction, the tension in the rope (T) is equal to the net force.

5. Tension in the rope (T) = Net Force (F_net)
T = F_net
T = 144 N

Therefore, the tension acting in the rope is 144 N.

To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

First, let's draw a free-body diagram of the crate on an inclined plane:

|\
| \
| \
| \
|____\

||||\
---------\

|

In this diagram, the gravitational force (mg) acting vertically downward can be split into two components: one parallel to the inclined plane (mg*sinθ) and one perpendicular to the inclined plane (mg*cosθ).

By using Newton's second law, we can write the equation:

Net force = m * acceleration

The net force acting on the crate is the vector sum of the gravitational force (mg*sinθ) and the tension force in the rope (T).

Along the y-axis:
Fy = mg * cosθ - N = 0
Therefore, the normal force (N) is equal to mg * cosθ.

Along the x-axis:
Fx = m * acceleration = T - mg * sinθ

Now, let's solve the magnitudes of the normal force and the tension force in the rope.

1. Magnitude of the normal force (N):
N = mg * cosθ
= (45 kg) * (9.8 m/s^2) * cos(35°)
≈ 380.3 N

Therefore, the magnitude of the normal force acting on the crate is approximately 380.3 N.

2. Magnitude of the tension force in the rope (T):
T - mg * sinθ = m * acceleration
T = m * acceleration + mg * sinθ
= (45 kg) * (3.2 m/s^-2) + (45 kg) * (9.8 m/s^2) * sin(35°)
≈ 504.0 N

Therefore, the magnitude of the tension force acting in the rope is approximately 504.0 N.

Net force= Tension + normal force + weigh

I used net force= ma.
I got Normal force= ma + mgCos(angle)
Normal force to be 505N.

Tension= ma + mgSin(angle)
= 396
Is this correct? or is the Net force = o.