A boy pulls his toy on a smooth horizontal surface with a rope inclined at 60'.the horizontal surface is 5N.CALCULATE
Tension equals t
T equals f
5cos60 degrees
5 multiple by ½
F equals 2.5N
this makes no sense to me. <<the horizontal surface is 5N>>
To calculate the tension force in the rope when a boy pulls his toy on a smooth horizontal surface with a rope inclined at 60 degrees, we can use trigonometry and decompose the forces into their components.
Let's denote the tension force in the rope as T and the weight of the toy as W. The weight can be calculated by multiplying the mass of the toy (m) by the acceleration due to gravity (g).
The weight of the toy is given by W = m * g.
To find the tension force in the rope, we need to first determine the vertical component of the weight. Since the surface is smooth and horizontal, there is no vertical acceleration acting on the toy. Therefore, the vertical component of the weight is balanced by an equal and opposite force from the surface. Thus, the vertical component of the weight is zero.
Next, we need to calculate the horizontal component of the weight. The horizontal component is given by W * sin(θ), where θ is the angle of inclination (60 degrees in this case).
Therefore, the horizontal component of the weight is W * sin(60).
Knowing that the weight is 5 N, we can calculate the horizontal component of the weight.
Horizontal component = 5 N * sin(60) = 5 * 0.866 = 4.33 N (approximately)
Since the rope is being pulled horizontally, the tension force in the rope is equal to the horizontal component of the weight.
Therefore, the tension force in the rope is approximately 4.33 N.