Net External Force on Crate: Magnitude & Direction

[anglum]

Homework Statement

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?

 

[anglum]

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]

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]

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]

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?