Why Does a Graduation Tassel Deflect Backward When a Car Accelerates?

[godkills]

Your friend's 12.4 g graduation tassel hangs on a string from his rearview mirror. (a) When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car. Explain. (b) If the tassel hangs at an angle of 6.44 degrees relative to the vertical, what is the acceleration of the card?


I am trying to understand the mechanics of forces yet some reason I don't understand it maybe it is the wording or my english trying to understand "deflects backward". Also I don't know how to figure out part b

The book says

When the car accelerates from the stoplight, the string must exert a forward force on the tassel in order to accelerate it in the horizontal direction at the same rate as the car's acceleration. The string must also continue to exert an upward force on the tassel to balance the force of gravity. As a result the tassel hangs at an angle, deflected toward the back of the car.

Somehow that isn't really getting to my head.

also part b answer is 1.11 m/s^2 i figured out the normal force and weight but i figured since the tassle isn't on the ground there is only normal force meaning the tension and no weight? Or there is always weight even if it is above ground?

 

[collinsmark]

The book says

When the car accelerates from the stoplight, the string must exert a forward force on the tassel in order to accelerate it in the horizontal direction at the same rate as the car's acceleration. The string must also continue to exert an upward force on the tassel to balance the force of gravity. As a result the tassel hangs at an angle, deflected toward the back of the car.

Somehow that isn't really getting to my head.

also part b answer is 1.11 m/s^2 i figured out the normal force and weight but i figured since the tassle isn't on the ground there is only normal force meaning the tension and no weight? Or there is always weight even if it is above ground?

Draw a free body diagram (FBD). This is the first step, and it is an important one.

Model the tassel as a massless string, attached to a ball of mass m at the end. (It turns out that you don't really even need to know the specific value of the mass. So just call it m.)

There are two forces acting on mass m: The tension T (from the string), and the force of gravity mg.

From Newton's second law, the vector sum of these forces must equal ma.

ma = ∑ F_i

(where boldface represents vectors.) Solve for a.

Hint: You already know the direction of the acceleration a. That piece of information is important.

Another hint: When summing the forces, you need to sum them as vectors! You'll need to break up the tension T into its horizontal and vertical components.

 

[tiny-tim]

hi godkills!

… since the tassle isn't on the ground there is only normal force meaning the tension and no weight? Or there is always weight even if it is above ground?

there is always weight

if the body is on the ground, then there is also a normal force (which is often, but not always equal and opposite to the weight)

I am trying to understand the mechanics of forces yet some reason I don't understand it maybe it is the wording or my english trying to understand "deflects backward".

"deflects" simply means that the angle is changed