How Do You Solve These Classic Physics Vector Problems?

[shayaan_musta]

1)What displacement must be added to a 50cm displacement in the +x-direction to give resultant displacement of 85cm at 25°?

2)A child is holding a wagon from rolling straight back down a driveway that is inclined at 20°to the horizontal. If the wagon 150N, with what force must the child pull on the handle if the handle is parallel to the incline?

3)A boat is propelled so as to travel with a speed of 0.50m/s in still water, moves directly across a river i.e. 60m wide. The river flows with a speed of 0.30m/s. (a) At what angle relative to straight-across, direction must the boat be pointed? (b) How long does it take to cross the river?

I have these 3 question. I don't know how solve it. Because I didn't get idea. If you people just at least give me idea, I will very great full to you.
Thanks a lot.

 

[BobG]

You should read this post explaining the guidelines: https://www.physicsforums.com/showthread.php?t=94379

But, to at least get you started, find the x and y components of the 85 cm vector. Then the rest of the question should be fairly straight forward.

Try to draw a picture of your problem. It makes it easier to figure out what you need to do to solve the problem.

 

[shayaan_musta]

You should read this post explaining the guidelines: https://www.physicsforums.com/showthread.php?t=94379

But, to at least get you started, find the x and y components of the 85 cm vector. Then the rest of the question should be fairly straight forward.

Try to draw a picture of your problem. It makes it easier to figure out what you need to do to solve the problem.

Ok I got your point. Thanks. Actually I am new to PF. Here is detail for what I tried to solve.
1) let x be the displacement which is to be added to 50cm in +x-direction to give the resultant of 85.
So, the equation can be 85=x+50 i.e. x=35cm. But my book says its 45cm at 53°. And also in answer how to use θ? What is wrong with my solution.

2)Here is a solution that I tried:
sinθ=perp/hyp
sin(20°)=150/F
where, F=force that is along inclined plane and to be find out.
F=150/sin(20°)
F=438.570N
But according to my textbook it should be 51N. How?

 

[BobG]

Ok I got your point. Thanks. Actually I am new to PF. Here is detail for what I tried to solve.
1) let x be the displacement which is to be added to 50cm in +x-direction to give the resultant of 85.
So, the equation can be 85=x+50 i.e. x=35cm. But my book says its 45cm at 53°. And also in answer how to use θ? What is wrong with my solution.

2)Here is a solution that I tried:
sinθ=perp/hyp
sin(20°)=150/F
where, F=force that is along inclined plane and to be find out.
F=150/sin(20°)
F=438.570N
But according to my textbook it should be 51N. How?

1) This one is worded strangely. The way I would read it is the resulting vector is 85 cm long with 25 degree angle to the x axis. You calculate the x component of the 85 cm vector (the 85 cm is the hypotenuse). Then you subtract the x-component from 50.

You assumed the x-component was 85 (i.e. the adjacent side was 85) and I don't think that's correct.

Edit: Actually, given the answer you provided, I now understand what they were asking. You need the x and y components of the 85cm vector. The vector you're calculating has the same y component, but it's x component is 50 cm less than the 85 cm vector's x-component.

2) Right approach, but your 150 is the hyp and the perp is parellel to the incline, mainly because of the direction you chose for the principal direction.

If you chose down the inclined plane to be the principal direction, then perp would be the force pushing perpendicular to the inclined plane and you'd use the cosine of 70 degrees to solve for the parallel component and still get the same answer since 70 and 20 are complementary angles.

Like I said, it helps to draw these out and compare the downward direction of the force to the inclinded plane. Either way works as good as the other - it's just a matter of which is easier to visualize when you draw it out.

Personally, I would tend to look at the drawing, figure out which direction will have the most force, and then pick the sine or cosine based on which will give the most sensible answer. Then occasionally double check against the definitions given in the book.

 

[shayaan_musta]

Sorry my internet was making trouble therefore I am late in reply to your answer.

1) This one is worded strangely. The way I would read it is the resulting vector is 85 cm long with 25 degree angle to the x axis. You calculate the x component of the 85 cm vector (the 85 cm is the hypotenuse). Then you subtract the x-component from 50.

You assumed the x-component was 85 (i.e. the adjacent side was 85) and I don't think that's correct.

Edit: Actually, given the answer you provided, I now understand what they were asking. You need the x and y components of the 85cm vector. The vector you're calculating has the same y component, but it's x component is 50 cm less than the 85 cm vector's x-component

Ok ok. Its really helping step that you told me. Thanks. I got your point. But
In the question, i.e.
What displacement must be added to a 50cm displacement in the +x-direction to give the resultant displacement of 85cm at 25°?
the statement;

what displacement must be added to a 50cm displacement in the +X-DIRECTION

How will you read this statement?
Means is it given for 50cm in the +x-direction? Or it is asking to add displacement in the +x-direction?
Which one is right?