Girl pulling a 6.5 kg sled at 35 degrees with friction

[runningirl]

Homework Statement

A girl is pulling a 6.5 kg sled by a rope. The rope make a 35 degree angle with the horizontal. Friction provides a resistive force of 15 N.

a) If she pulls with a force of 70 N, what will be the acceleration of the sled?

b)If she pulls with a force of 140 N, what happens differently?

Homework Equations

f=ma=net force-normal force

The Attempt at a Solution

0=70sin35+Fn-(6.5*9.8)
Fn=23.5 N

then i got tripped up...

 

[cdotter]

How did you get "0=70sin35+Fn-(6.5*9.8)"?

 

[runningirl]

it's the net force.

 

[cdotter]

Sum of the forces = ma.

There are two forces. The force pulling the sled and the resistive frictional force. You need to find the horizontal component of the pulling force (check your trig function.) The sum of the forces is simply those two added together. The mass is given. Solve for acceleration.

 

[runningirl]

Sum of the forces = ma.

There are two forces. The force pulling the sled and the resistive frictional force. You need to find the horizontal component of the pulling force (check your trig function.) The sum of the forces is simply those two added together. The mass is given. Solve for acceleration.

is the resistive force 15Fn?

 

[cdotter]

is the resistive force 15Fn?

Yes.

 

[runningirl]

coefficient of friction*Fn=Force of friction
coefficient*23.5=15
coefficient=.64 N

so -.64(23.5)+70(cos35)
=-15+60.62
75.62=(6.5)(a)
a=9.33 m/s/s

 

[cdotter]

You don't need to calculate the coefficient of friction (and you can't, from the data given.) Also, the coefficient of friction has no units.

The resistive frictional force is given and opposes the horizontal pulling force like so:

Fhorizontal=70N*cos(35)
Fresistive=-15N
ΣF=Fhorizontal+Fresistive=(70N*cos(35))+(-15N)=42.34N

Mass is given as m=6.5kg. Solve for acceleration (a.)

ΣF=ma
42.34N=(6.5kg)a
a=6.51 m/s^2