Home
/
Physics
/
Unless otherwise indicated, whenever a numerical value of gis required, take Answer ALL questions. g=9.8ms^-2 and give your answer to either 2 significant figures or 3 significant figures. A beam AB is supported by two vertical ropes, which are attached to the beam at points P and Q,where AP=0.3 m and BQ=0.3m The beam is modelled as a uniform rod, of length 2 m and mass 20 kg. The ropes are modelled as light inextensible strings. A gymnast of mass 50 kg hangs on the beam between P and Q. The gymnast is modelled as a particle attached to the beam at the point X, where PX =xm,0lt xlt 1.4 as shown in the diagram above. The beam rests in equilibrium in a horizontal position. (a) Show that the tension in the rope attached to the beam at P is (588-350x)N (3) (b) Find, in terms of x, the tension in the rope attached to the beam at Q. (c) Hence find,justifying your answer carefully, the range of values of the tension which could occur in each rope. Given that the tension in the rope attached at Q is three times the tension in the rope (d) find the value of x. (3)

Question

Unless otherwise indicated, whenever a numerical value of gis required, take Answer ALL questions. g=9.8ms^-2 and give your answer to either 2 significant figures or 3 significant figures. A beam AB is supported by two vertical ropes, which are attached to the beam at points P and Q,where AP=0.3 m and BQ=0.3m The beam is modelled as a uniform rod, of length 2 m and mass 20 kg. The ropes are modelled as light inextensible strings. A gymnast of mass 50 kg hangs on the beam between P and Q. The gymnast is modelled as a particle attached to the beam at the point X, where PX =xm,0lt xlt 1.4 as shown in the diagram above. The beam rests in equilibrium in a horizontal position. (a) Show that the tension in the rope attached to the beam at P is (588-350x)N (3) (b) Find, in terms of x, the tension in the rope attached to the beam at Q. (c) Hence find,justifying your answer carefully, the range of values of the tension which could occur in each rope. Given that the tension in the rope attached at Q is three times the tension in the rope (d) find the value of x. (3)

Unless otherwise indicated, whenever a numerical value of gis required, take Answer ALL questions. g=9.8ms^-2 and give your answer to either 2 significant figures or 3 significant figures. A beam AB is supported by two vertical ropes, which are attached to the beam at points P and Q,where AP=0.3 m and BQ=0.3m The beam is modelled as a uniform rod, of length 2 m and mass 20 kg. The ropes are modelled as light inextensible strings. A gymnast of mass 50 kg hangs on the beam between P and Q. The gymnast is modelled as a particle attached to the beam at the point X, where PX =xm,0lt xlt 1.4 as shown in the diagram above. The beam rests in equilibrium in a horizontal position. (a) Show that the tension in the rope attached to the beam at P is (588-350x)N (3) (b) Find, in terms of x, the tension in the rope attached to the beam at Q. (c) Hence find,justifying your answer carefully, the range of values of the tension which could occur in each rope. Given that the tension in the rope attached at Q is three times the tension in the rope (d) find the value of x. (3)

expert verifiedVerification of experts

Answer

3.9219 Voting
avatar
MiaProfessional · Tutor for 6 years

Answer

# Explanation:<br /><br />## Step 1: <br />First, we need to understand that the total weight of the system (beam and gymnast) is acting downwards due to gravity. The weight of the beam is \(20 \times g = 196\) N and the weight of the gymnast is \(50 \times g = 490\) N. The total weight is \(196 + 490 = 686\) N.<br /><br />## Step 2:<br />Next, we consider the equilibrium of moments about point P. The forces acting on the beam are the weight of the beam, the weight of the gymnast, and the tension in the ropes at P and Q. The beam is in equilibrium, so the sum of the moments about any point is zero.<br /><br />### \(T_{Q} \times 2 = 196 \times 1 + 490 \times x\)<br />### \(T_{Q} = \frac{196 + 490x}{2}\)<br /><br />## Step 3:<br />The tension in the rope at P can be found by considering the equilibrium of vertical forces. The sum of the upward forces (tensions) must equal the total weight.<br /><br />### \(T_{P} + T_{Q} = 686\)<br />Substitute the expression for \(T_{Q}\) from step 2 into this equation to find \(T_{P}\).<br /><br />### \(T_{P} = 686 - \frac{196 + 490x}{2} = 588 - 245x\)<br /><br /># Answer: (a) \(T_{P} = 588 - 245x\) N<br /><br /># Explanation:<br /><br />## Step 4:<br />To find the tension in the rope at Q, we substitute the expression for \(T_{P}\) from part (a) into the equilibrium of vertical forces equation.<br /><br />### \(T_{Q} = 686 - T_{P} = 686 - (588 - 245x)\)<br /><br /># Answer: (b) \(T_{Q} = 98 + 245x\) N<br /><br /># Explanation:<br /><br />## Step 5:<br />The range of values of the tension in each rope depends on the position of the gymnast (x). As x varies from 0 to 1.4, we substitute these values into the expressions for \(T_{P}\) and \(T_{Q}\) to find the range of tensions.<br /><br /># Answer: (c) For \(T_{P}\), the range is 343 N to 588 N. For \(T_{Q}\), the range is 98 N to 343 N.<br /><br /># Explanation:<br /><br />## Step 6:<br />Given that the tension in the rope at Q is three times the tension in the rope at P, we can set up the equation \(T_{Q} = 3T_{P}\) and solve for x.<br /><br />### \(98 + 245x = 3(588 - 245x)\)<br /><br /># Answer: (d) \(x = 0.6\) m
Click to rate:

Hot Questions

More x