Question

You pull a suitcase along the floor by exerting 43N at an angle. The force of friction is 27 N and the suitcase moves at a constant speed. What angle does the handle make with the horizontal?

Answer 1

The angle of the force is $51.1^{\circ}$

Explanation:

To solve this problem, we can apply Newton's second law along the horizontal direction of motion of the suitcase:

$\sum F_x = ma_x$

where

$\sum F_x$ is the net force along the x-axis

m is the mass of the suitcase

$a_x$ is the acceleration along the x-axis

The suitcase is moving at constant speed, so the acceleration is zero:

$a_x=0$

Therefore the net force must also be zero:

$\sum F_x = 0$ (1)

We have two forces acting along the horizontal direction:

- The component of the push (forward) in the horizontal direction, $F cos \theta$, with

F = 43 N

$\theta$ = angle of the force with the horizontal

- The  force of friction, $F_f = 27 N$, backward

So the net force can be written as

$\sum F_x = F cos \theta - F_f$ (2)

Combining (1) and (2),

$F cos \theta - F_f = 0$

And so we can find the angle:

$\theta = cos^{-1}(\frac{F_f}{F})=cos^{-1}(\frac{27}{43})=51.1^{\circ}$

Learn more about Newton's second law:

brainly.com/question/3820012

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