Question

Two adults and a child want to push a wheeled cart, which will be starting from rest, in the direction marked xx in the figure (Figure 1). The two adults push with horizontal forces F1→, of magnitude 117 N , and F2→ of magnitude 125 N , as shown in the figure.

a- Find the magnitude of the smallest force that the child should exert. You can ignore the effects of friction.

b-Find the angle that the force makes with the +x-direction. Take angles measured counterclockwise from the +x-direction to be positive.

c-If the child exerts the minimum force found in part (A) and in part (B) , the cart is found to accelerate at 1.90 m/s2 m/s2 in the +x-direction. What is the weight of the cart?

Answer 1

The magnitude of the smallest force that the child should exert is 242 N. The angle that the force makes with the +x-direction is 0 degrees. The weight of the cart is equal to the force exerted by the child.

Step-by-step explanation:

To find the magnitude of the smallest force that the child should exert, we need to consider the forces acting on the cart. The two adults are pushing with forces of 117 N and 125 N in the horizontal direction. Since the cart is starting from rest, the net force acting on it must be zero. This means that the force exerted by the child should be equal in magnitude but opposite in direction to the sum of the forces exerted by the adults. Therefore, the magnitude of the smallest force that the child should exert is 117 N + 125 N = 242 N.

To find the angle that the force makes with the +x-direction, we can use trigonometry. Let's denote the angle as θ. We can calculate θ using the following formula:

θ = arctan(Fy/Fx)

Here, Fx is the sum of the x-component of the forces exerted by the adults, and Fy is the y-component of the force exerted by the child. Since the forces are acting in the horizontal direction, Fy = 0. Therefore, the angle θ is 0 degrees, meaning that the force exerted by the child is in the same direction as the +x-direction.

To find the weight of the cart, we need to use Newton's second law of motion, which states that force equals mass times acceleration (F = ma). In this case, the force is the net force exerted on the cart, which is the force exerted by the child. The acceleration is given as 1.90 m/s². Rearranging the equation, we can solve for the mass of the cart: m = F/a. Once we have the mass, we can calculate the weight of the cart using the formula weight = mass x gravitational acceleration. The weight of the cart is equal to the force exerted by the child.

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Answer 2

The magnitude of the smallest force that the child should exert is 242 N and the force exerted by the child is purely horizontal. The weight of the cart is equal to the force exerted by the child.

a) To find the magnitude of the smallest force that the child should exert, we need to determine the net force acting on the cart. The net force is the vector sum of the forces exerted by the two adults and the child. Since the cart is starting from rest, the net force must be equal to the mass of the cart multiplied by the acceleration. Ignoring friction, we have:

`F_(net)` = m * a

`F_(1) + F_(2) + F_(child)` = m * a

`F_{child` = m * a -
`F_(1)` -
`F_(2)`

where,

`F_(1)` = 117 N,
`F_(2)` = 125 N

So, the smallest force the child should exert is 117 N + 125 N = 242 N in the direction opposite to that exerted by the adults.

b) To find the angle that the force exerted by the child makes with the +x-direction, we can use trigonometry. The force exerted by the child can be broken down into horizontal and vertical components:

`F_(child(horizontal)) = F_(child) * cos (theta)`

`F_(child(vertical)) = F_(child) * sin (theta)`

Since the cart is moving horizontally, the vertical component of the force exerted by the child should be zero. Therefore, we have:

`F_(child(vertical))` = 0

`F_(child)` * sin(theta) = 0

sin(theta) = 0

theta = 0 degrees

Therefore, the force exerted by the child is purely horizontal.

c) If the child exerts the minimum force found in part (a) and (b) and the cart accelerates at 1.90 m/s
`s^(2)`, we can use Newton's second law to find the mass of the cart:

`F_(net)` = m * a

`F_(1) + F_(2) + F_(child)` = m * a

`F_{child` = m * a -
`F_(1)` -
`F_(2)`

= m * 1.90 - 117 - 125

Once we have the mass of the cart, we can calculate its weight using the formula:

Weight = mass * gravitational acceleration

The weight of the cart is equal to the force exerted by the child.