Question

The frame is made from uniform rod which has a mass p per unit length. A smooth recesser constrains the small rollers at A and B to travel horizontally. Force P is applied to the frame through a cable attached to an adjustable colla Determine the magnitudes and directions of the normal forces which act on the rollers if () h = 0.212, (b) h - 0.502, and (c) h - 0.85L. The forces will be positive if up, negative if down. Evaluate your results for p=1.5 kg/m, L= 445 mm, and P-59 N. What is the acceleration of the me in each case? L P L B Answers: (a) - 0.211: A- B N N, m/s 1 m/s² (b) h - 0.50L: A - B N (c) h 0.85: A B- NA m/s2 GO TUTORIAL Policy 2000-2020 John Wiley & Sons, Inc. All Rights Reserved. A Division of John Wiley & Sons, Inc. Version 4.24.19, 17328 30 AN 9

Answer 1

Given data:

$\rho$ = 1.5 kg/m

L = 445 mm = 0.445 m

P = 59 N

Solution:

Draw the free-body diagram of the frame.

L С P L G ma mg A B 1/2 NA NB

The mass of the frame is a product to the length of the frame and mass per unit length

m Total length*mass per unit length m = 4L *P

Substitute L = 0.445 m, $\rho$ = 1.5 kg/m in the above equation.

m = 4*0.445*1.5

m = 2.67 kg

Write the sum of the forces in x-dirextion.

CF = 0 0 P - ma P=ma

Here P is the force applied to the frame at point C, a is the acceleration of the frame.

Substitute P = 59 N, m = 2.76 kg in the above equation.

59=2.67*a

a = 22.1 m/s.

The acceleration of the frame will be equal in all cases as mass and force applied are constant.

Write the equilibrium equation of forces along the y-axis.

.........Eq. (1)

2 Fy = 0 NA + NB - mg = 0 Where g = 9.81 m/s NA + NB - 2.67 * 9.81 = 0 NA + NB = 26.19

(a) h = 0.21L

Take a moment about point B.

MB=0 L Na + L +P + h – mg * 5 2 L ma = 0 NA * L +P+0.21L - mg * L 2 ma * L 2 = 0 1 1 NA + P* 0.21 – mg* ma * = 0 2 2 1 1 NA +59 * 0.21 – 2.67 * 9.81* 2.67 * 22 * = 0 2 2 NA = 30.08 N

Substitute the value of in Eq. (1)

NA = 30.08 N

30.08 + NB = 26.16 NB = -3.92 N

Therefore normal force acting at roller A is 30.08 N (Upwards)

and normal force acting at roller B is 3.92 N (Downwards)

(b) h = 0.5L

Take a moment about point B.

MB=0 L Na + L +P + h – mg * 5 2 L ma* 0 NA * L + P* 0.5L - mg* L 2 ma * L 2 = 0 1 NA + P* 0.5 – mg* =0 NA +59 * 0.5 – 2.67 * 9.81* 2.67 * 22* 0 2 NA = 12.97 N

Substitute the value of in Eq. (1)

NA = 12.97 N

12.97 + NB = 26.16 NB = 13.17 N

Therefore normal force acting at roller A is 12.97 N (Upwards)

and normal force acting at roller B is 13.17 N (Upwards)

(c) h = 0.85L

Take a moment about point B.

MB=0 L Na + L +P + h – mg * 5 2 L ma = 0 NA * L + P* 0.85L - mg * L 2 ma * L 2 = 0 1 NA + P* 0.85 – mg* 1 2 ma * = 0 2 1 1 NA +59 * 0.85 – 2.67 * 9.81* 2.67 * 22 * = 0 2 2 NA = -7.68 N

Substitute the value of in Eq. (1)

NA -7.68 N

7.68 + NB = 26.16 NB = 33.84 N

Therefore normal force acting at roller A is -7.68 N (downwards)

and normal force acting at roller B is 133.84 N (Upwards)

Answers.

(a) h = 0.21L: A=	30.08	N,  B=	-3.92	N, a =	22.1	m/s2
(b) h = 0.50L: A=	12.97	N,  B=	13.17	N, a =	22.1	m/s2
(c) h = 0.85L: A=	-7.68	N,  B=	33.84	N, a =	22.1	m/s2