Force and Tension in Cargo Box and Strap System

[minimax]

Hi! I have some physics problems here that hopefully sNomeone can help me on. Here's my questions and work. It's my first time on here, so hopefully I put enough information out there to clarify things. :D They are both cargo box/strap questions.

Homework Statement

1) A cargo box is pulled by a strap. The box is 80kg while the ange of the strap being pulled is 45degrees to the horizontal.

2) A cargo box is pulled by a strap:
Find the force (F) when
a) the normal force (N)=mg at 0 degrees, N is 0 at 30 degrees
b N=mg/2 at 30degrees and N=mg at 60 degrees
c)N=mg/2 at 90 degrees and N=mg at zero degrees

Homework Equations

I think it'll be easier if i put relevant equations with my work below

The Attempt at a Solution

1) For question one, I saw one that's sort of like mine except that there was friction.
I know that the vertical component is Tsin45=mg-N and I'm trying to find N.

I'm not sure about the horizontal component as there is not friction listed although realistically friction would be present and that for this question
Tx=Tcos45degrees only

2)
so far I've worked on section c
so i know:
Fytot=Wy+Fy+Ny=0
Fy=-Wy-Ny
Fy=-mg-N
Fysin90=-mg-(mg/2)
Fy=(-3/2)mg
Fx=Fcos90=0

F=sqrt(((-3/2)mg)^2)
so F=(3/2)mg??
or would it be mg/2?

Does this look like the right process?

Thank You very much for taking the time to help me!

 

[minimax]

ok, so I've done part c of the second question:

components:
Fy=-Wy(0,-1)-Ny(0,1)
so
Fy=mg-Ny
Fysin90=mg-(mg/2)
=mg/2

Fx=Fcos90=0

so F=sqrt((mg/2)^2)

therefore, F=mg/2

for the 30 degree question
i have so far:
mg=0 degrees
N=mg/2=30 degrees

Fysin30=mg-(mg/2)
(1/2)Fy=mg/2
Fy=mg

Fx=Fcos30(sqrt(3/2))

does this look right so far? (also still stuck on que 1)

 

[minimax]

Oh no! i just realized that i forgot the question for 1.

A cargo box is pulled by a strap. The box is 80kg while the angle of the strap being pulled is 45degrees to the horizontal. What is force N? (normal force)

 

[latitude]

Hey, does it say for question one it's on a frictionless surface? Or no?

 

[andrevdh]

The Attempt at a Solution

1) For question one, I saw one that's sort of like mine except that there was friction.
I know that the vertical component is Tsin45=mg-N and I'm trying to find N.

I'm not sure about the horizontal component as there is not friction listed although realistically friction would be present and that for this question
Tx=Tcos45degrees only

Correct. The problem do not say anything about the motion does it (eg. constant velocity)? If not then the horizontal component cannot be solved and you can only get N in terms of T from your vertical analysis.

 

[andrevdh]

2)
so far I've worked on section c
so i know:
Fytot=Wy+Fy+Ny=0
Fy=-Wy-Ny
Fy=-mg-N
Fysin90=-mg-(mg/2)
Fy=(-3/2)mg
Fx=Fcos90=0

F=sqrt(((-3/2)mg)^2)
so F=(3/2)mg??
or would it be mg/2?

Does this look like the right process?

Thank You very much for taking the time to help me!

When the strap is pulling upwards (ninety degrees) we have that

$$N + F = W$$

therefore

$$F = W - N$$

giving

$$F = mg - \frac{mg}{2}$$

for the 0 degree case one need more info about the motion (if friction is present and the motion is progressing at a constant velocity then one can again solve for F).