# TBR Physics Question

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#### ilovemedi

##### Full Member
10+ Year Member
This refers to Chapter 2, Passage IV, Q#25. You have three blocks on a flat floor with a rope connected to each one. (see pick below).

Question: If the blocks is pulled w/ constant velocity, which way should a person connect the rope to the block so the horizontal component of tension in the rope is the LEAST?

A)I
B)II
C)III
D) II and IIII

I know T-Ff=0... Thus, T=Ff... Wouldn't the horizontal compnent of tension in the rope be the LEAST in A) Block I since T = Ff/cos30, making it large. Thus II and III would make a larger horizontal component?

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If friction was the same, I, II and III should have the same horizontal component of the tension - it's exactly the opposite of the friction, since the block is moving with a constant velocity. The thing is, the tension from the rope in II and III has also a vertical component which changes the normal force on the block and correspondingly, the friction.

With that taken into account, the least horizontal tension is achieved in the case with the least friction, or with the least normal force. That case is B where the vertical component of the tension acts in an opposite direction to the weight of the block.

Btw, your logic applies to the magnitude of the tension on rope, not just to the horizontal component - that will be the same as Ff.

T - F(friction) = 0
T = F(friction)
T = N&#956;

For II, N = mg - Tsin&#952;. For III, N = mg + Tsin&#952;. Thus, N for II < N for III, indicating a smaller fictional force. Since the horizontal component of the tension must be equal to the fictional force to achieve constant velocity, II will allow this to be done with least amount of force.

If friction was the same, I, II and III should have the same horizontal component of the tension - it's exactly the opposite of the friction, since the block is moving with a constant velocity. The thing is, the tension from the rope in II and III has also a vertical component which changes the normal force on the block and correspondingly, the friction.

With that taken into account, the least horizontal tension is achieved in the case with the least friction, or with the least normal force. That case is B where the vertical component of the tension acts in an opposite direction to the weight of the block.

Btw, your logic applies to the magnitude of the tension on rope, not just to the horizontal component - that will be the same as Ff.

Well said.

OP, I just took the exam that you're referring to. A question like this should be solved using intuition before equations are recruited. If it isn't, time is wasted that could have otherwise been used on more mathematically rigorous questions.

Imagine pulling the box yourself &#8211; will you have an easier time pulling it if the rope is angled towards the ground? Probably not, since you're increasing the normal force, N. That rules out III. Since I is between II and III and is perfectly horizontal, I is ruled at as well.

Yes, the answer is B. Thanks milski and Endure. I completely forgot that N is affected by the vertical component. Solved everything. Thanks again!

Also - Endure, thanks for the insight. Will use intuition for future problems... I definitely don't want to waste time.