Net external forces


by anglum
Tags: external, forces
anglum
anglum is offline
#1
Sep25-07, 05:45 PM
P: 275
1. The problem statement, all variables and given/known data

A crate is puled to the right with a force of 75.3N and to the left with a force of 126.2N
it is pulled upward with a force of 652.2N and downward with a force of 231.9N

A - what is the net external force in the x direction

B- what is the net external force in the y direction

C - what is the magnitude of the net external force on the crate?

D - what is the direction of the net external force on the crate (measured from the positive x-axis witth counterclockwise positive) answer in units of degrees?


A - 75.3N-126.2N = -50.9N or 50.9N to the left??

B - 652.2N - 231.9N = 420.3N or 420.3N up???

c - the magnitude is the then 420.3N up and 50.9N to the left correct?

D - not sure how to do this part?????
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anglum
anglum is offline
#2
Sep25-07, 05:54 PM
P: 275
for c do i use the pythagorean theorem????

and for d do i use sin or cos based off of my answer for c to find the angle of the force?
hage567
hage567 is offline
#3
Sep25-07, 05:59 PM
HW Helper
P: 1,542
The components of the force in each direction form a right angle triangle. You can use the Pythagorean theorem to find the magnitude of the resultant vector. You can also use trig to find the angle you need.

anglum
anglum is offline
#4
Sep25-07, 06:04 PM
P: 275

Net external forces


are my answers correct for A and B..

and then for C i would do

50.9 squared + 420.3 squared = X squared

x = 423.3708 would be the magnitude correct?

and then it is the sin of 420.3/423.3708 would give me the answer to D???
hage567
hage567 is offline
#5
Sep25-07, 06:22 PM
HW Helper
P: 1,542
Your answers for the first three parts look reasonable to me. You're on the right track for (d), the question asks you for the angle from the positive x-axis. Have you drawn out a diagram of the directions of the net forces? What angle does sin(420.3/423.3708) give you in that diagram?


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