Join IDNLearn.com and start getting the answers you've been searching for. Ask your questions and receive prompt, detailed answers from our experienced and knowledgeable community members.
Sagot :
To find the total normal force acting on the cart, we need to consider both the gravitational force and the vertical component of the applied force. Here’s a step-by-step solution:
1. Determine the mass of the cart:
The mass [tex]\( m \)[/tex] of the cart is given as [tex]\( 7.32 \)[/tex] kg.
2. Identify the force applied and its angle:
The force [tex]\( F \)[/tex] applied by the shopper is [tex]\( 14.7 \)[/tex] N, directed at [tex]\( -32.7^\circ \)[/tex] below the horizontal.
3. Calculate the gravitational force acting downward:
The gravitational force [tex]\( F_g \)[/tex] can be calculated using the relation:
[tex]\[ F_g = m \times g \][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( 9.81 \ \text{m/s}^2 \)[/tex]):
[tex]\[ F_g = 7.32 \ \text{kg} \times 9.81 \ \text{m/s}^2 = 71.8092 \ \text{N} \][/tex]
4. Determine the vertical component of the applied force:
Since the force is applied at an angle below the horizontal, we need the vertical component [tex]\( F_v \)[/tex]. The vertical component can be found using the sine of the angle [tex]\( \theta \)[/tex]:
[tex]\[ F_v = F \times \sin(\theta) \][/tex]
where [tex]\( \theta = -32.7^\circ \)[/tex]. Converting this angle to radians:
[tex]\[ \theta_{\text{rad}} = -32.7^\circ \times \frac{\pi}{180} = -0.5707226654021458 \ \text{radians} \][/tex]
Now, calculate the vertical component:
[tex]\[ F_v = 14.7 \ \text{N} \times \sin(-0.5707226654021458) = 14.7 \ \text{N} \times (-0.5707226654021458) = -7.941532711021528 \ \text{N} \][/tex]
5. Calculate the total normal force:
The normal force [tex]\( n \)[/tex] is the sum of the gravitational force and the vertical component of the applied force:
[tex]\[ n = F_g + F_v \][/tex]
Substituting the values:
[tex]\[ n = 71.8092 \ \text{N} + (-7.941532711021528 \ \text{N}) = 63.867667288978474 \ \text{N} \][/tex]
Thus, the total normal force acting on the cart is:
[tex]\[ n = 63.867667288978474 \ \text{N} \][/tex]
1. Determine the mass of the cart:
The mass [tex]\( m \)[/tex] of the cart is given as [tex]\( 7.32 \)[/tex] kg.
2. Identify the force applied and its angle:
The force [tex]\( F \)[/tex] applied by the shopper is [tex]\( 14.7 \)[/tex] N, directed at [tex]\( -32.7^\circ \)[/tex] below the horizontal.
3. Calculate the gravitational force acting downward:
The gravitational force [tex]\( F_g \)[/tex] can be calculated using the relation:
[tex]\[ F_g = m \times g \][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( 9.81 \ \text{m/s}^2 \)[/tex]):
[tex]\[ F_g = 7.32 \ \text{kg} \times 9.81 \ \text{m/s}^2 = 71.8092 \ \text{N} \][/tex]
4. Determine the vertical component of the applied force:
Since the force is applied at an angle below the horizontal, we need the vertical component [tex]\( F_v \)[/tex]. The vertical component can be found using the sine of the angle [tex]\( \theta \)[/tex]:
[tex]\[ F_v = F \times \sin(\theta) \][/tex]
where [tex]\( \theta = -32.7^\circ \)[/tex]. Converting this angle to radians:
[tex]\[ \theta_{\text{rad}} = -32.7^\circ \times \frac{\pi}{180} = -0.5707226654021458 \ \text{radians} \][/tex]
Now, calculate the vertical component:
[tex]\[ F_v = 14.7 \ \text{N} \times \sin(-0.5707226654021458) = 14.7 \ \text{N} \times (-0.5707226654021458) = -7.941532711021528 \ \text{N} \][/tex]
5. Calculate the total normal force:
The normal force [tex]\( n \)[/tex] is the sum of the gravitational force and the vertical component of the applied force:
[tex]\[ n = F_g + F_v \][/tex]
Substituting the values:
[tex]\[ n = 71.8092 \ \text{N} + (-7.941532711021528 \ \text{N}) = 63.867667288978474 \ \text{N} \][/tex]
Thus, the total normal force acting on the cart is:
[tex]\[ n = 63.867667288978474 \ \text{N} \][/tex]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.