Solve Magnitude & Distance: 390 N North, 180 N East, m=270 kg & 10 N, m=30 kg

[willingtolearn]

[SOLVED] Magnitude - Distance

#1
The force of the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration.
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390 N North, 180 N East.
m = 270 kg

I never know how to find the magnitude, can some one help or explain the magnitude to me ?


#2
A shopper in a supermarket pushes a loaded cart with a horizontal force of 10 N. The cart has a mass of 30 kg. HOw far will it move in 3.0 s, starting from the rest? (Ignore friction)
-----------------------------------
Horizontal F = 10 N
m = 30kg
t = 3 s
d = rt
v (initial) = 0

I only got to this point ?
Help please ? Thanks

 

[Kurdt]

Well for part one you simply use pythagoras' theorem. Think of the components of force as two sides of a right angled triangle. For part two you will need to use one of the kinematic equations. Are you familiar with these?

They can be found here: https://www.physicsforums.com/showthread.php?t=110015

 

[willingtolearn]

#1
So the magnitude is the same as the resultant !
#2
I only know to the Newton 2nd Law

 

[Kurdt]

A resultant vector is a sum of vectors, the magnitude of the vector is the length of the vector.

For number two try applying the following kinematic equation.

s = ut+\frac{1}{2}at^2

,where s is the distance traveled and u is the initial velocity.

 

[willingtolearn]

#1
mag = 429.5 N
acc is NorthEast with the value 1.6 m/s2

#2
OH, i forgot that equation ! Forgot to think of the term displacement !
d = 1.5 m
Is that correct ?

 

[Kurdt]

You may need to work out an angle for number 1. I'm not sure if they'll let you get away with just north east. Other than that you're correct.

 

[willingtolearn]

IK ! Thanks