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#### caveman721

##### Full Member
7+ Year Member

please help, it seems I am stuck. If you could find the angle and show/explain the process that would be great.

A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling force can be applied to the box in two ways. It can point horizontally, or it can point above the horizontal at an angle 0 (theta). Wwhen the pulling force is applied horizontally, the kinetic frictional force acting on the box is twice as large as when it is applied at an angle 0 (theta). what is the value 0 (theta)?

please show me the process and explain, thanks again.

#### Pisiform

##### Oh Crap!!!
10+ Year Member
I am not sure but I will give it a shot:

When the force of 110N was horizontal:

F(k) = (mew)N
= (mew) 150

When the force of 110N was acting at an angle:

F(k) = (mew)N
= (mew)(150 - 110Sin(theta) )

Normal force in the second case would be less because you are subtracting the vertical component of the force acting at an angle theta.

F(k) horizontal is twice as large as F(k) at theta

(mew) 150 / 2 = (mew) (150 - 110Sintheta)

75 = 150 - 110Sin theta (mew cancels out)
theta = 42.9 degrees

#### cartman1980

##### Full Member
F(k) = (mew)N
= (mew)(150 - 110Sin(theta) )

wouldn't there also be a cos(theta) component acting against the horizontal frictional force?

#### Pisiform

##### Oh Crap!!!
10+ Year Member
wouldn't there also be a cos(theta) component acting against the horizontal frictional force?

Yes there will be, but it wouldn't affect or change your normal force as normal force is in vertical plane

PS: somebody correct me if I am wrong

#### ilovemcat

##### Full Member
Removed
In either case, the force applied is 110 N. That doesn't change. What changes then?

The problem tells you when the force applied is perfectly horizontal - the friction is "twice as great." But what is friction?

Friction = mk x Fnormal

Looking at this equation - you know that mk is a constant. Therefore, the only way the friction could be "twice as great" was if the normal force was twice as large.
So, if the normal force is twice as large (when the applied force is perfectly horizontal), it must be half as large when held at it's at an angle.

Let's consider only the vertical forces acting on the object:

You have the downward weight of the object.
You have the upward applied force.
You have the normal force (which we know needs to equal 75 N)
We also know the object isn't accelerating upwards or downwards, so Fnet = 0 N.

Let's set up the equation:

Fnet = 0N = Weight - Fapplied + Normal Force

Normal Force = Weight - Fapplied
75 N = 150 N - 110sin(theta) <== Note: "sin(theta)" because we're only considering the vertical component
110sin(theta) = 150N - 75N
110sin(theta) = 75N
sin(theta) = 75N/110N
sin(theta) = 0.68

So the angle "theta" must equal 42.98 or ~45 degrees if you had to estimate for the MCAT.