Angle of a force when towing a suitcase

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A woman is towing a 20kg suitcase at a constant speed with a 35N force on a strap at an unknown angle from the horizontal, while facing a 20N frictional force. The angle of the strap to the horizontal is calculated to be 55.2 degrees using the equation Tcos(θ) = F. The normal force exerted by the ground on the suitcase is determined to be approximately 167.3N, derived from the vertical force balance equation. The discussion highlights the application of vector components to solve for both the angle and the normal force. The solutions provided are useful for similar physics problems involving forces and motion.
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hi guys
ive got a problem where a woman is towing a 20kg suitcase at constant speed. She is pulling on a strap of unknown degrees from the horizontal. She pulls on the strap with a 35N force and the frictional force on the suitcase is 20N.
It asks for the angle that strap makes to the horizontal as well as the normal force the ground exerts on the suitcase.
Any clues
Thanx
 
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ok guys, I've worked it out
ill post my solutions just incase anyone was interested or you have a similar question

for part a
It says its moving at constant velocity, therefore to overcome the 20N friction there must be a 20N force in the horizontal direction, therefore this allows us to use reverse cos(20/35). This gives an angle of 55.2 degrees.

part b

use component method in vectors to find the vertical force of the strap, which is 35sin(55.2). minus this from mg( which is equal to N) and this gets 167N
 
barney007 said:
hi guys
ive got a problem where a woman is towing a 20kg suitcase at constant speed. She is pulling on a strap of unknown degrees from the horizontal. She pulls on the strap with a 35N force and the frictional force on the suitcase is 20N.
It asks for the angle that strap makes to the horizontal as well as the normal force the ground exerts on the suitcase.
Any clues
Thanx

Let the unknown angle=@

H: Tcos@-F=0

=>Tcos@=F

=>35cos@=20

=>cos@=(20/35)=(4/7)

=>@=arccos(4/7)=55.2deg

V: Tsin@+N-mg=0

=>N=mg-Tsin@

=>N=20g-35sin55.2=167.3N

Dirac.
 
Dirac said:
Let the unknown angle=@

H: Tcos@-F=0

=>Tcos@=F

=>35cos@=20

=>cos@=(20/35)=(4/7)

=>@=arccos(4/7)=55.2deg

V: Tsin@+N-mg=0

=>N=mg-Tsin@

=>N=20g-35sin55.2=167.3N

Dirac.

Did you not read his second post? He worked out the solutions already..
 
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