Finding Acceleration of 16.5kg Suitcase Given Forces

[shawonna23]

A passenger is pulling on the strap of a 16.5-kg suitcase with a force of 61.0 N. The strap makes an angle of 34.5° above the horizontal. A 37.8-N friction force opposes the motion (horizontal) of the suitcase. Determine the acceleration of the suitcase.

F applied= 61.0
F gravity= 16.5 x 9.8=161.7
F friction= 37.8
Normal Force= ?

Would the normal force = mg cos(34.5)

After I have all the forces, what do I do to determine the acceleration?

 

[Curious3141]

Why do you even need the normal force ? You're given the exact frictional force acting on the suitcase, you don't need to compute it using the coefficient of friction here.

Ignore all components acting vertically. What is the horizontal component of the applied force (the one pulling the bag) ? Given the opposing frictional force, what is the net horizontal force on the bag ? Using a well known relationship between force, mass and acceleration, what is the acceleration of the suitcase ?

 

[shawonna23]

How would I calculate the horizontal net force that is opposed by the frictional force?

Would it be F net = 16.5 X 9.8= 161.7

 

[dextercioby]

Why...?Is gravity on the horizontal direction...?

Daniel.

 

[shawonna23]

no, gravity isn't on the horizontal axis.

So, would I just add 61.0 N + 37.8N to get the Net horizontal force

 

[dextercioby]

No,u need to consider the fact that the force acts at an angle...

Daniel.

 

[Curious3141]

To the original poster, draw a force diagram and tell us what forces act on the suitcase.

Then tell us what are the horizontal components and the vertical components.

This will be good for you, because you seem to be shooting in the dark right now. You need to learn the correct concepts, not just blindly apply standard formulae.

 

[shawonna23]

Is this correct or did i use the wrong force?

a= 61.0cos(34.5) - 37.8 divided by 16.5 kg
a=0.756m/s2

 

[Curious3141]

The above is correct.