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Chapter 11: Problem 23

A \(1000-\mathrm{N}\) crate rests on a horizontal floor. It is being pulled up by two vertical ropes. The left rope has a tension of \(400 \mathrm{~N}\). Assuming the crate does not leave the floor, what can you say about the tension in the right rope?

Short Answer

Expert verified
Answer: The tension in the right rope is 600 N.

Step by step solution

01

Define the forces acting on the crate

The forces acting on the crate in the vertical direction are: the tension in the left rope (\(T_{left}=400 \mathrm{~N}\)), the tension in the right rope (\(T_{right}\)), and the weight of the crate (\(W=1000 \mathrm{~N}\)). All these forces are acting vertically, either upwards (tension forces) or downwards (weight).
02

Apply Newton's second law of motion

According to Newton's second law of motion, the net force acting on an object is equal to the mass of the object multiplied by its acceleration (\(F_{net} = m \times a\)). In this case, the acceleration of the crate is zero, since it does not leave the floor. Thus, the net force acting on the crate in the vertical direction is also zero. So, we can write the equation for the net force as: \(F_{net} = T_{left} + T_{right} - W = 0\)
03

Solve the equation for the tension in the right rope

We will now solve the equation from Step 2 for the unknown tension in the right rope (\(T_{right}\)): \(T_{left} + T_{right} - W = 0 \) Plugging in the given values for \(T_{left}\) and \(W\), we get: \(400 \mathrm{~N} + T_{right} - 1000 \mathrm{~N} = 0\) Now, we will solve for \(T_{right}\): \(T_{right} = 1000 \mathrm{~N} - 400 \mathrm{~N} = 600 \mathrm{~N}\)
04

State the result

The tension in the right rope is \(600 \mathrm{~N}\). This means that the right rope has higher tension than the left rope, and both ropes combined provide enough upward force to counteract the weight of the crate, ensuring it does not leave the floor.

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Most popular questions from this chapter

A ladder of mass \(M\) and length \(L=4.00 \mathrm{~m}\) is on a level floor leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is \(\mu_{\mathrm{s}}=\) 0.600 , while the friction between the ladder and the wall is negligible. The ladder is at an angle of \(\theta=50.0^{\circ}\) above the horizontal. A man of mass \(3 M\) starts to climb the ladder. To what distance up the ladder can the man climb before the ladder starts to slip on the floor?

A track has a height that is a function of horizontal position \(x\), given by \(h(x)=x^{3}+3 x^{2}-24 x+16\). Find all the positions on the track where a marble will remain where it is placed. What kind of equilibrium exists at each of these positions?

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You have a meter stick that balances at the \(55-\mathrm{cm}\) mark. Is your meter stick homogeneous?

If the wind is blowing strongly from the east, stable equilibrium for an open umbrella is achieved if its shaft points west. Why is it relatively easy to hold the umbrella directly into the wind (in this case, easterly) but very difficult to hold it perpendicular to the wind?
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