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## Precalculus

### Course: Precalculus>Unit 6

Lesson 7: Vector components from magnitude and direction

# Vector components from magnitude & direction: word problem

Given the magnitudes and directions of the forces two people are applying to a box, find the overall force that is applied in the direction of the target. Created by Sal Khan.

## Want to join the conversation?

• I am not convinced with sal's explanation and this is how i feel. Please check what is wrong with this reasoning.

If we plot same scenario on 2D graph assuming box is at origin 0, 0 and target is at some positive number on y axis then :
Vertical component of vector a becomes cos(35) * 330 = 270 (approximated) and horizontal coponent become sin(35) * 330 = 189.
Since horizontal component is acting away from origin in negative direction, we can represent vector a = -189i + 270j.

Similarly vertical component of vector b becomes cos(15) * 300 = 290 (approximated) and horizontal coponent become sin(15) * 300 = 78.
Since horizontal component is acting away from origin in positive direction, we can represent vector b= 78i + 290j.

Note that vertical components of both vectors are acting towards target which we have assumed is in positive direction, so they are positive.

Hence resultant vector of and b can be given as a + b = (-189+78)i + (270+290)j = -111i + 560j.

So resultant vector is not going in the direction of target, rather it is moving away from target in negative direction horizontally.

To calaculate magnitude of resultant vector call it c.
c = sqrt(-111*-111 + 560*560) = 570.89 Newton

What is wrong with it ? •  The problem is that we are only concerned with the the vectors in the direction of the target (perhaps the box is on rails), so all we are concerned with are the 270 and 290 force vectors and we can consider the remaining force (the i component in your case) as wasted energy.
• I know that the result is the same, but if we have considered the dashed line as our horizontal axis, wouldn't vector b have an angle of -15? • Yes, but cos(-15) = cos(15).

Whats more interesting to me is that there is a force that is perpendicular to the line between the box and the target which could result in the box never reaching the target. Person A exerts a perpendicular force which is 330sin(35)= 189N and person B exerts a perpendicular force 300sin(-15)= -78N. Notice these two are in opposite directions. Since the perpendicular force of A is higher than the perpendicular force of B, we know the box will never reach the target. It will veer off to the left of the target.

So how much does person A need to back off on pushing so they exactly cancel out the components perpendicular to the line of motion? If we change person A force from 330N to F and we want all the forces perpendicular to the desired line of motion to be zero, we get Fsin(35) + 300sin(-15) = 0 or F = 135.4N. So the moral of the story is, if Person A wants to push at a angle more off course from target then person B, dont push so hard. Let person B do most of the work :)
• What is the whole "sub x" thing for? • a "sub x" means the amount of force in the x direction from vector a

b "sub x" means the amount of force in the x direction from vector b

Although this video didn't dwell on it, a "sub y" would be the amount (magnitude) of force in the y direction from vector a and so on
• Would the box really move towards the target exactly? Or it might end up at some other spot? • It will definitely end up at some other spot! The problem Sal is interested in is not where the box will end up, or in which direction it moves, but rather: what forces are exerted on the box in the precise direction of the box . This comes down to calculating the component of each force in this specific direction. The final question is: who exerts more force directly toward the box ? (This question is asked at around .)
• Will the box move in the given direction ( if it is not moving in that direction currently ) ? • How do I find the magnitudes of two vectors, a, b, if I'm only given just the angles for a and b, and also given the magnitude of a third vector, c, s.t. a+b+c=0 ?

I've been running in circles trying to figure this out.. • If you apply forces A and B simultaneously to the box, would it end up in the same place as you would get when you apply A and then B? • Yes, think of it this way, force A exerts 20 Newtons of force on the box, then force B exerts a negative 30 Newtons of force on the box. Lets say every 10 Newtons the box travels 1 meter. So if the box at first was at the position of 0 meters, the position now would be negative 1 meter (because 20 newtons [force A] plus negative 30 newtons [force B] is equal to negative 10 newtons as its total or net force). So converting the net force into the length the box traveled, that would be negative one meter. The same exact thing occurs when you apply the forces simultaneously. The twenty newtons in force A cancels with the negative 20 newtons in force B, leaving just the negative 10 newtons in force B. So the answers are both the same.
(1 vote)
• Why does Sal use a "sub x" instead of ||a "sub x"|| at ? Is there a specific benefit to not using vector notation or is it simply a matter of preference? • According to the parallelogram law of addition of two vectors:
vector R = vector A + vector B
=> ||vector R|| = sqrt(A^2 + B^2 +2AB cos theta ) [ where theta = the angle bet. vec. A and vec. B]
But, as per this formula the answer comes 571.115 N, why this difference ?  