Vectors and Forces Questions

[hyorinmaru]

Homework Statement

#1: Vector A has a length of 2.0 meters and lies at an angle of 30 degrees from the x-axis. Vector B has a length of 5.0 meters and lies at an angle of 120 degrees from the x-axis.
What is C=B-A (x and y components) and what is the magnitude of C?

#2: What are the solutions to the following equation:
x^2+7x-3=0

#3: How do you tell which direction an object accelerates when there are 3 different directions given?

#4:You are in charge of a large catapult which is attempting to hurl a large stone up to a castle 15 m above your current location. You set up the catapult a distance x from the edge of a cliff. You know that the stone will be launched with a speed of 40 m/s and at an angle of 45 degrees. At what distance x from the castle should you place the catapult for it to hit the castle on its downward trajectory?

#5: What forces act on a 50 kg block which is being pushed across a frictionless floor (from rest) with a constant force of 10 Newtons at an angle of 30 degrees from the horizontal? What is the order of the strengths and the directions? After being pushed for 10 seconds how fast will the box be travelling? How far will it have traveled?

#6: With what amount of force must a person pull on the vertical portion of a rope to make a 1000 kg refrigerator travel up a 45 degree rough (0.10) plane at constant speed? When the refrigerator just about reaches the top of the inclined plane the rope breaks. With what magnitude of acceleration does the refrigerator accelerate down the ramp?

Homework Equations

I'm not sure which acceleration formula to use...

The Attempt at a Solution

#1: 90 degrees, 3 meters
#2: x=-7+/- sqrt(61) over 2
#3: not sure
#4:
#5:
#6

 

[Fallinleave]

There are 5 equations:
V^2=V₀^2 + 2a(x-x₀)
V=V₀+at
$$\Delta$$x=V₀t+(1/2)at^2
$$\Delta$$x=Vt-(1/2)at^2
$$\Delta$$x=(1/2)t(V+V₀)

V: velocity at time t. V₀: initial velocity. a: acceleration. t: time

All you need to do is list what is known from the question and choose one of the equation that contains all the known variable, then you just solve for the unknown.
there are many cases that you have to solve two equations to get the final answer.

 

[Fallinleave]

sorry for posting another one because I just realized that I haven't talked about how to deal with force.

For force question, please do the followings:
1. Draw free body diagram
2. choose axis(which way is positive x direction and which way is positive y direction)
3. write Newton's second law equations for the each axis on each diagram.
4. look back to the question on what we trying to find, and solve the equation.

Thanks

 

[hyorinmaru]

I don't really understand how to use the formulas for #4, 5 and 6

 

[rwisz]

:

#1: Vector C can be found by subtracting the x and y components of vector A from vector B. The x and y components of vector A are 2.0*cos(30) and 2.0*sin(30) respectively. The x and y components of vector B are 5.0*cos(120) and 5.0*sin(120) respectively. Therefore, the x component of C is 5.0*cos(120) - 2.0*cos(30) = -0.5 meters and the y component of C is 5.0*sin(120) - 2.0*sin(30) = 2.598 meters. The magnitude of C can be found using the Pythagorean theorem: sqrt((-0.5)^2 + (2.598)^2) = 2.65 meters.

#2: The solutions to the equation can be found using the quadratic formula: x = (-b +/- sqrt(b^2 - 4ac))/2a. In this case, a=1, b=7, and c=-3. Plugging in these values gives x = (-7 +/- sqrt(61))/2. Therefore, the solutions are x = (-7 + sqrt(61))/2 and x = (-7 - sqrt(61))/2.

#3: The direction of acceleration can be determined by looking at the net force acting on the object. If the net force is in the same direction as the velocity, the object will accelerate in that direction. If the net force is in the opposite direction of the velocity, the object will decelerate. If the net force is perpendicular to the velocity, the object will change direction but not speed.

#4: The distance x can be found using the projectile motion equations. The horizontal distance traveled by the stone is given by x = v*cos(theta)*t. The vertical distance traveled is given by y = v*sin(theta)*t - (1/2)*g*t^2. Since the stone will be at the same height when it hits the castle as it was when it was launched, the vertical distance traveled should be equal to the height of the castle. Therefore, setting y=15 and solving for t gives t = 2.04 seconds. Plugging this value into the equation for x gives x = 40*cos(45)*2.04 = 40*sqrt(2). Therefore, the catapult