Skip to main content
Ch 05: Applying Newton's Laws
Chapter 5, Problem 5

A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.)

Verified step by step guidance
1
Identify the forces acting on the picture frame: the weight of the frame acting downward (W), and the tensions in the two wires (T) acting upward and outward at an angle to the vertical.
Set up the equilibrium conditions for the forces. Since the frame is in equilibrium, the sum of the vertical forces must equal zero and the sum of the horizontal forces must also equal zero.
Express the vertical and horizontal components of the tension in terms of the angle \( \theta \) and the tension T. The vertical component of each tension is \( T \cos(\theta) \) and the horizontal component is \( T \sin(\theta) \).
Write the equations for the equilibrium conditions: For vertical forces, \( 2T \cos(\theta) = W \) and for horizontal forces, the forces should cancel out, implying \( 2T \sin(\theta) = 0 \) which simplifies to \( T \sin(\theta) = T \sin(\theta) \).
Solve the vertical equilibrium equation for \( \theta \) using the given that each tension is 0.75 times the weight of the frame, i.e., \( T = 0.75W \). Substitute T in the vertical equilibrium equation and solve for \( \theta \).

Verified Solution

Video duration:
0m:0s
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tension in Wires

Tension is the force exerted along a wire or rope when it is pulled tight by forces acting from opposite ends. In this scenario, the tension in each wire must balance the weight of the picture frame, which is the force due to gravity acting on it. Understanding how tension works is crucial for analyzing the forces acting on the frame.
Recommended video:
Guided course
09:57
Magnetic Force on Current-Carrying Wire

Equilibrium of Forces

An object is in equilibrium when the net force acting on it is zero. For the picture frame, this means that the vertical components of the tension in the wires must equal the weight of the frame, while the horizontal components must cancel each other out. This concept is essential for determining the angles at which the wires are positioned.
Recommended video:

Trigonometric Relationships

Trigonometric functions relate the angles of a triangle to the lengths of its sides. In this problem, the angles formed by the wires with the vertical can be analyzed using sine and cosine functions. These relationships help in calculating the vertical and horizontal components of the tension forces, which are necessary for solving the problem.
Recommended video:
Guided course
3:26
Trigonometry
Related Practice
Textbook Question
In a laboratory experiment on friction, a 135-N block resting on a rough horizontal table is pulled by a horizontal wire. The pull gradually increases until the block begins to move and continues to increase thereafter. Figure E5.26 shows a graph of the friction force on this block as a function of the pull. (a) Identify the regions of the graph where static friction and kinetic friction occur.

1244
views
Textbook Question
Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate A has mass mA, and crate B has mass mB. The coefficient of kinetic friction between each crate and the surface is μk. The crates are pulled to the right at constant velocity by a horizontal force F Draw one or more free-body diagrams to calculate the following in terms of mA, mB, and μk: (b) the tension in the rope connecting the blocks.

915
views
Textbook Question

In another version of the 'Giant Swing' (see Exercise 5.50), the seat is connected to two cables, one of which is horizontal (Fig. E5.51). The seat swings in a horizontal circle at a rate of 28.0 rpm (rev/min). If the seat weighs 255 N and an 825-N person is sitting in it, find the tension in each cable.

2979
views
1
rank
1
comments
Textbook Question
When jumping straight up from a crouched position, an average person can reach a maximum height of about 60 cm. During the jump, the person's body from the knees up typically rises a distance of around 50 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (c) In terms of this jumper's weight w, what force does the ground exert on him or her during the jump?
280
views
Textbook Question
A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (c) Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is 1.62 m/s2. (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?
308
views
Textbook Question
An 8.00-kg block of ice, released from rest at the top of a 1.50-m-long frictionless ramp, slides downhill, reaching a speed of 2.50 m/s at the bottom. (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.0 N parallel to the surface of the ramp?
452
views