Physics question?

Question

A passenger is pulling on the strap of a 15.0kg suitcase with a force of 70.0N. The strap makes an angle of 35 degrees above the horizontal. A 37.8N friction force opposes the horizontal motion of the suitcase. Determine the acceleration of the suitcase.

I'm okay with most force questions, but I have no idea what to do with the angle.

Answer 1

The angle just allows us to split the force applied through the strap into vertical and horizontal components.

vertical component = 70 * sin(35)
horizontal component = 70 * cos(35)

we can assume the vertical forces are balanced so that just leaves the horizontal

overall horizontal force = force from the strap - friction force
overall horizontal force = 70 * cos(35) - 37.8 = 19.54 N

then get accelerations using
acceleration = force/mass
acceleration 19.54/15 = 1.3 m s^-2

Answer 2

1.30 meters per second squared.

What you do with the angle is multiply the pulling force by the cosine of the angle to get the actual force pulling in the horizontal direction.

Net force = mass (acceleration)

Now, the net force consists of two terms:
Pulling = 70cos(35)
Friction = -37.8
Net force = 70cos(35) - 37.8 = 19.5406

So a = F/m = 19.5406 / 15 = 1.3027

The answer should actually be rounded to three significant figures

a = 1.30 m/(s^2) :)

Answer 3

The horizontal component of 70 N is 70 cos 35.
This is the pulling force. This is opposed by 37.8 N
Resultant pulling force = 70 cos 35 - 37.8 = 19.54 N
Acceleration = Force / mass = 19.54 / 15 = 1.3 m/s^2

Answer 4

Determine the resultant force, you can get the angle and the amount of force. The amount of force is equal to the acceleration of the suitcase.

Answer 5

The force is the hypotenuse of a right triangle.
The weight is the vertical side, the friction, horizontal.
Got it?

Answer 6

you know that we have :F cos θ - f = ma
so: 70 *cos 35 - 37.8=15 a

a=(70*cos 35 - 37.8)/15