- The 160-lb crate is supported by cables AB, AC, and AD.Determine the tension in these wires. Determine the tension developed in cable AB. Determine the tension developed in cable AC. C.) Determine the tension developed in cable.
- A man drags a 150-lb crate across a floor by pulling on a rope inclined 15 degrees above the horizontal. A) If the coefficient of static friction is 0.50, what tension in the rope is required to start the crate moving? The answer is supposedly 68 lb. I start by assuming the normal force = weight, so Fn = 150 lb.
- Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 400 lb sin60 sin40 sin80 TT AC BC °° ° (a) 400 lb (sin60 ) AC sin80 T = ° ° 352 lbT AC = (b) 400 lb.

PROBLEM 2.62 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 300 lb in cable AC and 150 lb in cable BC, determine (a) the magnitude of the largest force P which may be applied at C, (b) the corresponding value of.

## The 150 Lb Crate Is Supported By Cables Made

Oh I knew I'd have to come crawling onto the internet in search of help, help me physics friends !!

**1. Determine the tension in each of the ropes holding the object. Rope T1 is at an angle of 60 degrees, rope T2 is at an angle of 55 degrees, the suspended block has a mass of 300kg.****2. w = mg****3. 300kg * 9.8 m/s = 2940 N, or the weight of the block, but since this I've just divided the weight by the sin of each rope's given angle. Completely baffled and very tired, looking for a push in the right direction so my brain can worky again.**

The answers are:

T1 = 1862 N , T2 = 1620 N but I have no clue how to arrive at themThe answers are:

T1 = 1862 N , T2 = 1620 N but I have no clue how to arrive at them

## The 150 Lb Crate Is Supported By Cables Per

A man drags a 150-lb crate across a floor by pulling on a rope inclined 15 degrees above the horizontal. a) If the coefficient of static friction is 0.50, what tension in the rope is required to start the crate moving?

The answer is supposedly 68 lb.

I start by assuming the normal force = weight, so Fn = 150 lb. The force of friction operating against the direction of motion is Ff = Fn*coefficient of static friction, which is at its maximum.

So 150*.5 = 75 lb = Ff

Now the x component of the force required to move the crate should be equal to the frictional force (Ff). So Cos 15 = 75/Frope, and Frope turns out to be 77.65 lb. This is wrong, but oddly it is 9.8 off from the correct answer (though I realize this is irrelevant since the problem is in lbs). What am I doing wrong? Thanks

The answer is supposedly 68 lb.

I start by assuming the normal force = weight, so Fn = 150 lb. The force of friction operating against the direction of motion is Ff = Fn*coefficient of static friction, which is at its maximum.

So 150*.5 = 75 lb = Ff

Now the x component of the force required to move the crate should be equal to the frictional force (Ff). So Cos 15 = 75/Frope, and Frope turns out to be 77.65 lb. This is wrong, but oddly it is 9.8 off from the correct answer (though I realize this is irrelevant since the problem is in lbs). What am I doing wrong? Thanks