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When a boy pulls his sled with a rope, the rope makes an angle of 35° with the horizontal. If a pull of 16 pounds on the rope is needed to move the sled, what is the horizontal component force?
9 lb
13 lb
19 lb
22 lb

Respuesta :

Answer:

  13 lb

Step-by-step explanation:

The horizontal component of the force is the magnitude of the force multiplied by the cosine of the angle with the horizontal.

  (16 lb)·cos(35°) ≈ 13.1 lb ≈ 13 lb

Answer:

b.13 lb

Step-by-step explanation:

We are given that  a boy pulls his sled with a rope .

The rope makes an angle with horizontal =[tex]35^{\circ}[/tex]

If a pull on the rope is needed to move the sled=16 pounds

We have to find the horizontal component of force

We know that horizontal component force=[tex]fcos\alpha[/tex]

Therefore, we have f=16 pound

Then, horizontal component of force=[tex]16cos 35^{\circ}[/tex]

Horizontal component of force=[tex]16\times 0.819=13.1lb[/tex]

Hence, horizontal component of force=13 lb

Answer:b.13 lb