course phy 201 FINISHED!!!! 019. Vector quantities
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Given Solution: If we identify the easterly direction with the positive x axis then these movements correspond to the displacement vector with x component 3 miles and y component 4 miles. This vector will have length, determined by the Pythagorean Theorem, equal to `sqrt( (3 mi)^2 + (4 mi)^2 ) = 5 miles. The angle made by this vector with the positive x axis is arctan (4 miles/(three miles)) = 53 degrees. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q002. When analyzing the force is acting on an automobile as it coasts down the five degree incline, we use and x-y coordinate system with the x axis directed up the incline and the y axis perpendicular to the incline. On this 'tilted' set of axes, the 15,000 Newton weight of the car is represented by a vector acting straight downward. This puts the weight vector at an angle of 265 degrees as measured counterclockwise from the positive x axis. What are the x and y components of this weight vector? Question by student and instructor response:I have my tilted set of axes, but I cannot figure out how to graph the 15,000 N weight straight downward. Is it straight down the negative y-axis or straight down the incline? ** Neither. It is vertically downward. Since the x axis 'tilts' 5 degrees the angle between the x axis and the y axis is only 85 degrees, not 90 degrees. If we start with the x and y axes in the usual horizontal and vertical positions and rotate the axes in the counterclockwise direction until the x axis is 'tilted' 5 degrees above horizontal, then the angle between the positive x axis and the downward vertical direction will decrease from 270 deg to 265 deg. ** It might help also to magine trying to hold back a car on a hill. The force tending to accelerated the car down the hill is the component of the gravitational force which is parallel to the hill. This is the force you're trying to hold back. If the hill isn't too steep you can manage it. You couldn't hold back the entire weight of the car but you can hold back this component. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x component of the weight vector =15000 N* cos (265 degrees) = about -1300 N y component of the weight vector =15,000 N * sin(265 degrees) = about -14,900 N Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The x component of the weight vector is 15,000 Newtons * cos (265 degrees) = -1300 Newtons approximately. The y component of the weight vector is 15,000 Newtons * sin(265 degrees) = -14,900 Newtons. Note for future reference that it is the -1300 Newtons acting in the x direction, which is parallel to the incline, then tends to make the vehicle accelerate down the incline. The -14,900 Newtons in the y direction, which is perpendicular to the incline, tend to bend or compress the incline, which if the incline is sufficiently strong causes the incline to exert a force back in the opposite direction. This force supports the automobile on the incline. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q003. If in an attempt to move a heavy object resting on the origin of an x-y coordinate system I exert a force of 300 Newtons in the direction of the positive x axis while you exert a force of 400 Newtons in the direction of the negative y axis, then how much total force do we exert on the object and what is the direction of this force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Use Pathagoreon Theorem to find the magnitude of the total force. magnitude = `sqrt( (300 N)^2 + (-400 N)^2 ) = 500 N. The angle the force is in the counterclockwise direction beginning from the positive x axis and = arctan ( -400 N / (300 N) ) = -53 deg Because of this we have to show this as: -53 degrees + 360 degrees = 307 degrees. Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Force is a vector quantity, so we can determine the magnitude of the total force using the Pythagorean Theorem. The magnitude is `sqrt( (300 N)^2 + (-400 N)^2 ) = 500 N. The angle of this force has measured counterclockwise from the positive x axis is arctan ( -400 N / (300 N) ) = -53 deg, which we express as -53 degrees + 360 degrees = 307 degrees. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q004. If I exert a force of 200 Newtons an angle of 30 degrees, and you exert a force of 300 Newtons at an angle of 150 degrees, then how great will be our total force and what will be its direction? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My x component of force = 200 N * cosine (30 degrees) = about 173 N My y component of force = 200 N * sine (30 degrees) = 100 N The force I exert will have to equal to the 2 forces in the x and y directions. Your x component of force = 300 N * cosine (150 degrees) = -260 N Your y component of 300 N * sine (150 degrees) = 150 N The force you exert will also have to be equal to the 2 forces in the x and y directions. In the direction of x we will have a total force of -87 N In the direction of y we will have a total force of 250 N Total force: x component = -87 N y component = 250 N Use the Pathagoreon Theorem and angle of arctan to find the magnitude and direction: magnitude of force = `sqrt( (-87 Newtons) ^ 2 + (250 Newtons) ^ 2) = about 260 N angle force is directed (counterclockwise from positive x axis) = arctan (250 N/(-87 N) ) + 180 deg = -71 deg + 180 deg = 109 degrees Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: My force has an x component of 200 Newtons * cosine (30 degrees) = 173 Newtons approximately, and a y component of 200 Newtons * sine (30 degrees) = 100 Newtons. This means that the action of my force is completely equivalent to the action of two forces, one of 173 Newtons in the x direction and one of 100 Newtons in the y direction. Your force has an x component of 300 Newtons * cosine (150 degrees) = -260 Newtons and a y component of 300 Newtons * sine (150 degrees) = 150 Newtons. This means that the action of your force is completely equivalent the action of two forces, one of -260 Newtons in the x direction and one of 150 Newtons in the y direction. In the x direction and we therefore have forces of 173 Newtons and -260 Newtons, which add up to a total x force of -87 Newtons. In the y direction we have forces of 100 Newtons and 150 Newtons, which add up to a total y force of 250 Newtons. The total force therefore has x component -87 Newtons and y component 250 Newtons. We easily find the magnitude and direction of this force using the Pythagorean Theorem and the arctan. The magnitude of the force is `sqrt( (-87 Newtons) ^ 2 + (250 Newtons) ^ 2) = 260 Newtons, approximately. The angle at which the force is directed, as measured counterclockwise from the positive x axis, is arctan (250 Newtons/(-87 Newtons) ) + 180 deg = -71 deg + 180 deg = 109 deg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: ********************************************* Question: `q005. Two objects, the first with a momentum of 120 kg meters/second directed at angle 60 degrees and the second with a momentum of 80 kg meters/second directed at an angle of 330 degrees, both angles measured counterclockwise from the positive x axis, collide. What is the total momentum of the two objects before the collision? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1st object momentum’s x component = 120 kg m/s * cos (60 deg) = 60 kg m/s y component = 120 kg m/s * sine (60 deg) = 103 kg m/s 2nd object momentum’s x component = 80 kg m/s* cos (330 deg) = 70 kg m/s y component = 80 kg m/s * sine (330 degr) = -40 kg m/s. Total momentum’s x component = 60 kg m/s + 70 kg m/s = 130 kg m/s y component 103 kg m/s+ (-40 kg m/s) = 63 kg m/s Magnitude of total momentum = `sqrt((130 kg m/s) ^ 2 + (63 kg m/s) ^ 2) = about 145 kg m/s Direction of total momentum angle = arctan (63 kg m/s/ (130 kg m/s)) = about 27 degrees Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The momentum of the first object has x component 120 kg meters/second * cosine (60 degrees) = 60 kg meters/second and y component 120 kg meters/second * sine (60 degrees) = 103 kg meters/second. The momentum of the second object has x component 80 kg meters/second * cosine (330 degrees) = 70 kg meters/second and y component 80 kg meters/second * sine (330 degrees) = -40 kg meters/second. The total momentum therefore has x component 60 kg meters/second + 70 kg meters/second = 130 kg meters/second, and y component 103 kg meters/second + (-40 kg meters/second) = 63 kg meters/second. The magnitude of the total momentum is therefore `sqrt((130 kg meters/second) ^ 2 + (63 kg meters/second) ^ 2) = 145 kg meters/second, approximately. The direction of the total momentum makes angle arctan (63 kg meters/second / (130 kg meters/second)) = 27 degrees, approximately. "