A 33-kg bowling ball is held up on a frictionless vertical wall by a string.The string goes from the bowling ball to an attachment point on the wall directly above the ball.The string makes an angle of 34 $\circ$ with respect to vertical.The horizontal force that the wall exerts on the ball is
A)324 N.
B)480 N.
C)195 N.
D)521 N.
E)218 N.

Question

Quizplus · Accepted Answer

The horizontal force that the wall exerts on the ball is equal to the horizontal component of the tension in the string. To find this, we first need to calculate the tension in the string, which balances the weight of the ball. The weight of the ball is $mg = 33 \times 9.8 = 323.4 \, \text{N}$. The tension in the string has a vertical component that balances this weight, and it can be found using trigonometry, specifically the sine of the angle (since the angle given is with respect to the vertical, and we're looking for the force opposite to this angle, which supports the weight of the ball).Let $T$ be the tension in the string. The vertical component of the tension is $T \cos(\theta)$, where $\theta = 34^\circ$. So, we have $T \cos(34^\circ) = 323.4 \, \text{N}$. Solving for $T$ gives us $T = \frac{323.4}{\cos(34^\circ)}$.The horizontal force that the wall exerts on the ball is the horizontal component of the tension, which is $T \sin(34^\circ)$. Substituting the value of $T$ from above, we get the horizontal force as $\frac{323.4}{\cos(34^\circ)} \sin(34^\circ)$.Calculating this gives a value close to 218 N, which corresponds to option E.

A 33-Kg Bowling Ball Is Held Up on a Frictionless ∘\circ∘

A 33-Kg Bowling Ball Is Held Up on a Frictionless $\circ$