Recitation%2011%20ch%209

Recitation%2011%20ch%209 - PHY 2053 Recitation Week 12...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PHY 2053 Recitation Week # 12 Problem # Ch. 9 # 12,25,37,44,49,57,60 12. A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 450 N walk on the overhanging part of the plank before it just begins to tip? REASONING At every instant before the plank begins to tip, it is in equilibrium, and the net torque on it is zero: τ Σ = (Equation 9.2). When the person reaches the maximum distance x along the overhanging part, the plank is just about to rotate about the right support. At that instant, the plank loses contact with the left support, which consequently exerts no force on it. This leaves only three vertical forces acting on the plank: the weight W of the plank, the force F R due to the right support, and the force P due to the person (see the free-body diagram of the plank). The force F R acts at the right support, which we take as the axis, so its lever arm is zero. The lever arm for the force P is the distance x . Since counterclockwise is the positive direction, Equation 9.2 gives W W 0 or W W Px x P τ Σ =- = = l l (1) SOLUTION The weight W = 225 N of the plank is known, and the force P due to the person is equal to the person’s weight: P = 450 N. This is because the plank supports the person against the pull of gravity, and Newton’s third law tells us that the person and the plank exert forces of equal magnitude on each other. The plank’s weight W acts at the center of the uniform plank, so we have (see the drawing) 1 1 W W 2 2 or d L L d + = =- l l (2) where d = 1.1 m is the length of the overhanging part of the plank, and L = 5.0 m is the length of the entire plank.Substituting Equation (2) into Equation (1), we obtain ( 29 ( 29 ( 29 1 1 2 2 225 N 5.0 m 1.1m 0.70 m 450 N W L d x P -- = = = * 25. A 1220-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N crate hangs from the far end of the beam. Using the data shown in the x L/2 F R W P d W l Free-body diagram of the plank Axis drawing, find (a) the magnitude of the tension in the wire and (b) the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam. REASONING The drawing shows the beam and the five forces that act on it: the horizontal and vertical components S x and S y that the wall exerts on the left end of the beam, the weight W b of the beam, the force due to the weight W c of the crate, and the tension T in the cable. The beam is uniform, so its center of gravity is at the center of the beam, which is where its weight can be assumed to act. Since the beam is in equilibrium, the sum of the torques about any axis of rotation must be zero ( 29 τ Σ = , and the sum of the forces in the horizontal and vertical directions must be zero ( 29 0, x y F F Σ = Σ = . These three conditions allow us to determine the magnitudes of...
View Full Document

This note was uploaded on 09/21/2011 for the course PHY 1500 taught by Professor Staff during the Spring '09 term at University of Central Florida.

Page1 / 6

Recitation%2011%20ch%209 - PHY 2053 Recitation Week 12...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online