Question

Question 9 (1 point) Saved For the backstop as shown if the maximum pressure between lining and drum is 200 psi, belt width is 0. inch, the friction coefficient is 0.15, and drum has a radius of 9 inches, what is the engaging force corresponding to the maximum belt load? 30 lbf 540 lbf 270 lbf O 60 lbf

Answer 1

Given Information :

• Maximum pressure between the lining and the drum :
  Pi= 200 psi = 200 lb/in
  
• Friction coefficient :
  u= 0.15
  
• Drum radius :
  r = 9 in
  
• Belt width :
  w = 0,5 in
  

To Find :  Engaging force P corresponding to maximum belt load

Solution :

Let
$T_1=$ Tension on the tight side of the belt
$T_2=$ Tension on the slack side of the belt

From the belt friction equation we can derive the relation between the tensions.

ha I

Where
$\theta=$ Angle of wrap in radians.

In this case the angle of wrap is : $\theta=270^0=4.7123\ rad$

Formulating the realtionship :

ha I

se. 1544.7123

T = T, * 2.0275

The tension $T_1$ represents the tension on the tight side of the belt. Since the given sense of rotation is counter-clockwise, the moment created due to $T_1$ should be in clockwise direction. Hence the tension force $T_1$ should be perpendicular to the engaging lever.

From the given information about the maximum pressure we can compute the tension on the tight side

T1 = P*A

Where
$A=$ Area of contact $=w*r$

Computing the maximum tension:

T1 = P*A

Ti = 200 * 9*0.5 = 900 lb

Using this compute the tension on the slack side

T = T, * 2.0275

900 = T, * 2.0275

T2 = 443.8964 lb

Since we have both the tensions, now we need to take moments about the given hinge to compute the value of Engaging force P

$T_1*3=P*10$

Substitute the values :

$900*3=P*10$

P = 270 lbf

Hence the engaging force corresponding to the maximum belt load is  
P=270 lbf
.

Option 3 is the right choice.