A green car with a mass of 900 kg is placed in the back of a yellow
tow truck that has a pivot at one end. The car is held in place with a
chain at the top end of the lift, as pictured below. The chain holding
the car in place can withstand a maximum force of 3000 N.
a) If the tow truck lifts its bed to an angle of θ = 15◦
, will the chain snap? Assume that the car is in
neutral, so friction is negligible between the wheels of the car and the bed of the truck.
Later, the same tow truck picks up a minivan with a mass of 1500 kg. When the truck lifts its bed to an
angle of θ = 20◦
, the chain holding the minivan breaks.
b) With what acceleration does the minivan roll down the back of the truck? Assume that the minivan
is in neutral, so friction is negligible between the wheels of the minivan and the bed of the truck.
a) T=mg sin 15
=(900 kg)(9.8 m/s^2)(sin 15)
= 2282.78 N, T<3000 N
Therefore, the chain will not snap.
b) i)T=mg sin 20
= (1500 kg)(9.8 m/s^2)(sin 20)
= 5027.69 N, T>3000 N
Therefore, the chain will snap.
ii) a= g sin20
= (9.8 m/s^2)(sin 20)
= 3.35 m/s^2
The car is held in place with a
chain at the top end of the lift, as pictured below.>>
This statement is too vague for physics.
Thanks for asking this idk how to do this one either
To determine whether the chain will snap in the given scenario, we need to analyze the forces acting on the car when the tow truck lifts its bed to an angle of θ = 15° in part (a).
Let's start by drawing a free-body diagram of the car when it is at an angle of 15°.
__
/\ |
Car / \ |
(mass = 900 kg) / \ |
- - - - -
In this diagram, the force of gravity acting on the car (its weight) can be split into two components: one parallel to the inclined surface (car's weight along the slope) and the other perpendicular to the inclined surface. The perpendicular component is balanced by the normal force between the car and the truck bed, which equals the car's weight.
Now, the component of the car's weight parallel to the incline can be found using trigonometry. The angle θ = 15° is the same as the angle between the inclined surface and the horizontal.
We can use the equation: weight_parallel = weight * sin(θ)
Given that the car's mass is 900 kg, the acceleration due to gravity is approximately 9.8 m/s^2, and the angle θ is 15°, we can calculate:
weight_parallel = 900 kg * 9.8 m/s^2 * sin(15°)
Next, we need to compare this weight_parallel with the maximum force the chain can withstand, which is 3000 N. If weight_parallel is greater than 3000 N, the chain will snap; otherwise, it will not snap.
To solve part (a), you can calculate weight_parallel and compare it with 3000 N to determine whether the chain will snap.
Now let's move on to part (b) where we need to find the acceleration at which the minivan rolls down the back of the truck.
When the chain holding the minivan breaks, the only force that acts on it is the force of gravity (weight). This force can be split into two components, just like before: one along the incline and one perpendicular to it.
Since the minivan is in neutral and friction is negligible, there is no force opposing its motion and causing acceleration. In this case, the component of the minivan's weight parallel to the incline will be responsible for its acceleration down the back of the truck.
To find the component of the minivan's weight parallel to the incline, we use the equation: weight_parallel = weight * sin(θ)
Given that the mass of the minivan is 1500 kg, the acceleration due to gravity is approximately 9.8 m/s^2, and the angle θ is 20°, we can calculate:
weight_parallel = 1500 kg * 9.8 m/s^2 * sin(20°)
The calculated weight_parallel will be the net force acting on the minivan, and since force = mass * acceleration (F = m * a), we can set weight_parallel as the resultant force acting on the minivan and solve for its acceleration using the equation:
acceleration = weight_parallel / mass
Substituting the values, you can calculate the acceleration at which the minivan rolls down the back of the truck.