My question is about TBR in-chapter question 2.16a on a 2-pulley system and the applied force to support a mass. The picture below is similar to the one for this question; but, in TBR, the weight of the mass is

View attachment Picture 8.png

The question asks how much force must be applied (Fa) to support the mass (m) (assuming the roes and pulleys are massless and frictionless). The applied force is to the diagonal downward part of the rope (as in the picture above).

A. 2mg

B. mg

C. mg/2

D. mg/3

-Isolating the lower pulley, which is in equilibrium (i.e. zero acceleration): downward force of mg and upward force of 2T (tension T in 2 vertical parts of the rope); so, 2T = mg; T = mg/2

-If Fa just supports the mass then the applied force is just the tension; so, since T = mg/2, the applied force to the diagonal downward part of the rope is mg/2.

-Wouldn't using the tension in the vertical part of the rope give the y-components of tension for the diagonal downward part of the rope?

-Since the question is asking for the applied force/tension in the diagonal part of the rope, shouldn't the answer be:

Applied force is diagonal part of rope = T / sin(theta) (or T / cos(theta), depending on which angle is

Since, T = mg/2;

Doesn't TBR's answer assume that the angle theta is 90 degrees?

Another way to look at it (see picture):

Since the pulley system is in equilibrium, the downward force on the right side of the upper pulley cancels out with the upward force on the left side of the lower pulley. So, what cancels out the upward force on the right side of the lower pulley, is the y-components of the downward force on the left side of the upper pulley. So, wouldn't the applied force on the diagonal downward part of the rope need to take this y-component aspect into account?

*m*(rather than 100 kg), and the applied force to the diagonal downward part of the rope is*applied force (Fa)*(rather than 50 kg):View attachment Picture 8.png

The question asks how much force must be applied (Fa) to support the mass (m) (assuming the roes and pulleys are massless and frictionless). The applied force is to the diagonal downward part of the rope (as in the picture above).

__The options:__A. 2mg

B. mg

C. mg/2

D. mg/3

__Their correct answer:__C__Their explanation:__-Isolating the lower pulley, which is in equilibrium (i.e. zero acceleration): downward force of mg and upward force of 2T (tension T in 2 vertical parts of the rope); so, 2T = mg; T = mg/2

-If Fa just supports the mass then the applied force is just the tension; so, since T = mg/2, the applied force to the diagonal downward part of the rope is mg/2.

__What I don't understand:__-Wouldn't using the tension in the vertical part of the rope give the y-components of tension for the diagonal downward part of the rope?

-Since the question is asking for the applied force/tension in the diagonal part of the rope, shouldn't the answer be:

Applied force is diagonal part of rope = T / sin(theta) (or T / cos(theta), depending on which angle is

*theta*); let's use Fa = T / sin(theta)Since, T = mg/2;

**applied force = mg / 2sin(theta)**Doesn't TBR's answer assume that the angle theta is 90 degrees?

Another way to look at it (see picture):

Since the pulley system is in equilibrium, the downward force on the right side of the upper pulley cancels out with the upward force on the left side of the lower pulley. So, what cancels out the upward force on the right side of the lower pulley, is the y-components of the downward force on the left side of the upper pulley. So, wouldn't the applied force on the diagonal downward part of the rope need to take this y-component aspect into account?

Last edited: Jul 9, 2012