Physics 2006 JAMB Past Questions - Question 12: a mirror of weight 75n hung by a cord from a hook on a wall if the cord makes an angle of 30 with the horizontal,the tension in the cord is

Correct Answer

D

Explanation

for vertical equilibrium, we have 2Ty = 75NBut Ty= adj = cos 60 degree T hyptherefore Ty = T Cos 60 degree2TY= 2TCos 60 degree2x T x 0.5 = 75 T = 75Ntherefore total tension in the rope = 2T= 2 x 75 = 150NTo find the tension in the cord holding the mirror, we can break down the forces acting on it. The weight of the mirror can be split into two components: one parallel to the direction of the cord (tension), and the other perpendicular to it (vertical).

Given:
Weight of the mirror (force due to gravity) = 75 N
Angle between the cord and the horizontal = 30 degrees

First, calculate the vertical component of the weight (mg), where m is the mass of the mirror and g is the acceleration due to gravity (approximately 9.81 m/s²):

Vertical component of weight = mg * cos(30°)

Since the mass is not given, you need to calculate it. You can use the formula:

Weight = mass * gravity

So, mass = Weight / gravity
mass = 75 N / 9.81 m/s² ≈ 7.65 kg

Now, calculate the vertical component of the weight:

Vertical component of weight = 7.65 kg * 9.81 m/s² * cos(30°)

Vertical component of weight ≈ 7.65 kg * 9.81 m/s² * 0.866 (cos(30°)) ≈ 65.56 N

Now, the tension in the cord must balance this vertical component of the weight. So, the tension in the cord is also 65.56 N.

The tension in the cord is approximately 65.56 N when the cord makes a 30-degree angle with the horizontal, and the mirror has a weight of 75 N.