Question

Beth, a construction worker, attempts to pull a stake out of the ground by pulling on a rope that is attached to the stake. The rope makes an angle of 46.6 ◦ with the horizontal. If Beth exerts a force of 124 N on the rope, what is the magnitude of the upward component of the force acting on the stake? Answer in units of N.

Answer 1

Answer:

90.1 N

Explanation:

We are given that

The ropes makes an angle with horizontal=$46.6^{\circ}$

Beth exerts a force on the rope=124 N

We have to magnitude of the upward component of the force on the stake.

Vertical component is always sin and horizontal component is always cos.

Therefore, the magnitude of the upward component of the force acting on the stake=$124 sin 46.6^{\circ}$

The magnitude of the upward component of the force acting on the stake=$124 \times 0.727$

The magnitude of the upward component of the force acting on the stake=90.148 N

Hence, the magnitude of the upward component of force acting on the stake=90.1 N

Answer 2

Answer:

The upward component force in magnitude is 90.095 N

Solution:

As per the question:

The angle that the rope make with the horizontal, $\theta = 46.6^{\circ}$

The force exerted by Beth on the rope, F = 124 N

Now, the upward component of the force that acts on the stake is the vertical component or the cosine component of the force that Beth exerts on the rope.

The magnitude of this component of force is given by:

$F_{up} = Fsin\theta$

$F_{up} = 124\times sin46.6^{\circ} =90.095 N$

$F_{up} = 90.095 N$