A 1130 kg -car is being pulled up a frictionless ramp at a constant speed. The cable makes an angle of 31.0 degrees above the surface of the ramp, and the ramp itself rises at 25.0∘ above the horizontal. Find the tension in the cable. How hard does the surface of the ramp push on the car?
literally did not answer the quesitno damon thats not poggers L + ratio
5460N
Damon did answer both questions, dumb dumb
Well, I'm not really good with physics, but I'll give it a shot!
To find the tension in the cable, you need to consider the forces acting on the car. Since the car is being pulled up at a constant speed, the net force on the car is zero. So, the tension in the cable must equal the force of gravity pulling the car downward.
Now, to find the force of gravity, you need to consider the weight of the car. The weight of an object is given by the equation: weight = mass × acceleration due to gravity.
Given that the car has a mass of 1130 kg and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the weight of the car as follows:
weight = 1130 kg × 9.8 m/s² = 11,074 N.
Since the tension in the cable equals the weight of the car, the tension in the cable is 11,074 N.
As for the second part of your question, the surface of the ramp pushes on the car with a force equal in magnitude but opposite in direction to the weight of the car. So, the surface of the ramp pushes on the car with a force of 11,074 N downwards.
I hope that helps, and I apologize if my attempt at humor didn't quite hit the mark!
To find the tension in the cable, we can analyze the forces acting on the car along the ramp.
1. Weight Force (W): The weight of the car is given by the product of its mass (m) and the acceleration due to gravity (g = 9.8 m/s²). So, W = m * g.
2. Normal Force (N): The normal force is the force exerted by the surface of the ramp on the car perpendicular to the ramp's surface.
3. Tension Force (T): The tension in the cable pulls the car up the ramp.
For the car to be moving at a constant speed, the net force acting on it must be zero. This means that the vector sum of the forces along the ramp must add up to zero.
Now, let's break the weight force and the tension force into their components along the ramp and perpendicular to the ramp.
The weight force can be broken down into two components:
- Parallel to the ramp (W_parallel): W_parallel = W * sin(θ),
- Perpendicular to the ramp (W_perpendicular): W_perpendicular = W * cos(θ).
Similarly, we can break down the tension force into two components:
- Parallel to the ramp (T_parallel): T_parallel = T * cos(φ), where φ is the angle the cable makes with the horizontal.
- Perpendicular to the ramp (T_perpendicular): T_perpendicular = T * sin(φ).
Since the car is moving at a constant speed along the ramp, the net force along the ramp is zero. Therefore, the parallel components of the weight and tension must be equal:
W_parallel = T_parallel.
Now, let's calculate the values:
1. Calculate the weight force:
W = m * g,
W = (1130 kg) * (9.8 m/s²).
2. Calculate the parallel component of the weight force:
W_parallel = W * sin(θ),
W_parallel = (W) * sin(31.0°).
3. Set the parallel component of the weight force equal to the parallel component of the tension force:
W_parallel = T_parallel.
4. Solve for the tension force:
T_parallel = W_parallel.
To find how hard the surface of the ramp pushes on the car, we need to calculate the perpendicular component of the tension force.
5. Calculate the perpendicular component of the tension force:
T_perpendicular = T * sin(φ).
The force exerted by the surface of the ramp on the car is equal to the perpendicular component of the tension force (T_perpendicular).
component of weight down ramp
= 1130 * 9.81 sin 25
component of cable tension up ramp
= T cos 31
so
T cos 31 = 1130 * 9.81 * sin 25
Part B
weight component normal to ramp down
1130 * 9.81 * cos 25
reduce by component of T up normal to ramp
T sin 31
so
1130 * 9.81 * cos 25 - T sin 31