Tension of a pole attached to a wall at an angle

[Robert5742]

Homework Statement

A rod extends from a vertical wall at an angle of 27º from the horizontal. A 2.5-kg lamp is mounted at the end of the rod. After John throws his 300-g winter coat over the lamp, what is the magnitude of the tension in the rod?

The answers choices are:
A) 30N
B) 70N
C) 27N
D) 22N

Homework Equations

F = ma

The Attempt at a Solution

Since the only force acting on it is $F_t$ tension is equal to T = mg/sin(theta)

So when I plug in my knowns I get T = (9.8*2.8) (2.5kg + 300 g (300g = .3kg)/(sin(27)) which is equal to 28.691... which is not one of the answer choices. Am I using the wrong formula?

 

[Student100]

Homework Statement

A rod extends from a vertical wall at an angle of 27º from the horizontal. A 2.5-kg lamp is mounted at the end of the rod. After John throws his 300-g winter coat over the lamp, what is the magnitude of the tension in the rod?

The answers choices are:
A) 30N
B) 70N
C) 27N
D) 22N

Homework Equations

F = ma

The Attempt at a Solution

Since the only force acting on it is $F_t$ tension is equal to T = mg/sin(theta)

So when I plug in my knowns I get T = (9.8*2.8) (2.5kg + 300 g (300g = .3kg)/(sin(27)) which is equal to 28.691... which is not one of the answer choices. Am I using the wrong formula?

How are you getting ~29N, check your math?

$F_{Ty}$ is not the only force acting on the rod in the y direction as you stated. The sum of the forces is $Tsin(\theta)-(m_1+m_2)g = 0$

There are some problems with this line of thinking. Imagine the rod was mounted from the wall horizontally with respect to the ground, with some mass hanging from one end. Would the tension in the rod in this case be zero, so that we could hang any mass on the end without stressing the rod/mounting?

Was there more information in the problem? Or am I just too tired to see it. Have you studied torques? Using the above we could choose C, although I'm not happy with that.

 

[Robert5742]

How are you getting ~29N, check your math?

$F_{Ty}$ is not the only force acting on the rod in the y direction as you stated. The sum of the forces is $Tsin(\theta)-(m_1+m_2)g = 0$

There are some problems with this line of thinking. Imagine the rod was mounted from the wall horizontally with respect to the ground, with some mass hanging from one end. Would the tension in the rod in this case be zero, so that we could hang any mass on the end without stressing the rod/mounting?

Was there more information in the problem? Or am I just too tired to see it. Have you studied torques? Using the above we could choose C, although I'm not happy with that.

This is the entire problem. We haven't really gone into torque that much. I only know the equations but not really how to apply them. So Torque = Forces x Length, but the given question doesn't give Length, so how should I go about solving it? Also I got about 29 since 9.8*2.8 = 27.44/(sin(27)) is about 29.