Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
The uniform lamina L, shown shaded in Figure 2, is formed by removing the square PORV, of side 2a, and the square RSTU, of side 4a, from a uniform square lamina ABCD, of side 8a. The lines QRU and VRS are straight. The side AD is parallel to PV and the side AB is parallel to PQ. The distance between AD and PV is a and the distance between AB and PQ is a. The centre of mass of L is at the point G. 42 (a) Show that the distance of G from the side AD is a 11 (5) The mass of L is M. A particle of mass kM is attached to L at C. The lamina, with the attached particle, is freely suspended from B and hangs in equilibrium with BC making an angle of 45° with the horizontal. (b) Find the value of k.

The uniform lamina L, shown shaded in Figure 2, is formed by removing the square PORV, of side 2a, and the square RSTU, of side 4a, from a uniform square lamina ABCD, of side 8a. The lines QRU and VRS are straight. The side AD is parallel to PV and the side AB is parallel to PQ. The distance between AD and PV is a and the distance between AB and PQ is a. The centre of mass of L is at the point G. 42 (a) Show that the distance of G from the side AD is a 11 (5) The mass of L is M. A particle of mass kM is attached to L at C. The lamina, with the attached particle, is freely suspended from B and hangs in equilibrium with BC making an angle of 45° with the horizontal. (b) Find the value of k.

icon
Related questions
Question

b please

A P V 2a 2a U R 8a 4a 4a S B 8a T D Figure 2 The uniform lamina L, shown shaded in Figure 2, is formed by removing the square PORV, of side 2a, and the square RSTU, of side 4a, from a uniform square lamina ABCD, of side 8a. The lines QRU and VRS are straight. The side AD is parallel to PV and the side AB is parallel to PQ. The distance between AD and PV is a and the distance between AB and PQ is a. The centre of mass of L is at the point G. (a) Show that the distance of G from the side AD is a 42 11 (5) The mass of L is M. A particle of mass kM is attached to L at C. The lamina, with the attached particle, is freely suspended from B and hangs in equilibrium with BC making an angle of 45° with the horizontal. (b) Find the value of k. (A)
Transcribed Image Text:A P V 2a 2a U R 8a 4a 4a S B 8a T D Figure 2 The uniform lamina L, shown shaded in Figure 2, is formed by removing the square PORV, of side 2a, and the square RSTU, of side 4a, from a uniform square lamina ABCD, of side 8a. The lines QRU and VRS are straight. The side AD is parallel to PV and the side AB is parallel to PQ. The distance between AD and PV is a and the distance between AB and PQ is a. The centre of mass of L is at the point G. (a) Show that the distance of G from the side AD is a 42 11 (5) The mass of L is M. A particle of mass kM is attached to L at C. The lamina, with the attached particle, is freely suspended from B and hangs in equilibrium with BC making an angle of 45° with the horizontal. (b) Find the value of k. (A)
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer