Respuesta :
You want to draw a free body diagram of the forces on the sled in the horizontal x-direction.
If you visualize the system in an x-y coordinate plane, the force along the x-direction is the angle it makes with the x-axis multiples by the force.
The angle made with the x-axis is cosine of the angle theta.
Please see picture attached.
If you visualize the system in an x-y coordinate plane, the force along the x-direction is the angle it makes with the x-axis multiples by the force.
The angle made with the x-axis is cosine of the angle theta.
Please see picture attached.
When a force, F, is applied to an object at an angle,θ, to compute for the horizontal and vertical components, we have the following
[tex] F_{vertical} = Fsin(\theta) [/tex]
[tex] F_{horizontal} = Fcos(\theta) [/tex]
So, for this particular problem, given that the force applied to the sled is 4.00 N at 62⁰. Thus, the horizontal component of the force is equal to
F = 4.00(cos62⁰) = 1.88 N
Therefore, the horizontal force is 1.88 N.
Answer: 1.88 N
[tex] F_{vertical} = Fsin(\theta) [/tex]
[tex] F_{horizontal} = Fcos(\theta) [/tex]
So, for this particular problem, given that the force applied to the sled is 4.00 N at 62⁰. Thus, the horizontal component of the force is equal to
F = 4.00(cos62⁰) = 1.88 N
Therefore, the horizontal force is 1.88 N.
Answer: 1.88 N