A block is hung by a string from inside the roof of avan. When the van goes straight ahead at a speed of 24 m/s,the block hangs vertically down. But when the van maintains this same speed around an unbanked curve (radius = 175m) the block swings toward the outside of the curve, then the string makes an angle theta with the vertical. Find theta.

Question

Alix Ortiz · Accepted Answer

While the van is moving straight at constant velocity, thereis no acceleration, so it hangs straight down as it would at any constant velocity. When the van enters the curve, the block &quot;wants&quot; to keep going straight but the string pulls it in uniformcircular motion.  From the diagram, set it up using Newton&#039;s second law. For the x direction: F=Tsin⁡(θ)=mv2R (For circular motion, a=v2R) Now for the y direction: F=Tcos⁡θ−mg=0 or T=mgcos⁡θ Putting this in the first equation yields: mgcos⁡θsin⁡θ=mv2R or tan⁡θ=v2Rg Putting in the nunbers I get θ=18.55∘

Vasquez · Accepted Answer

Answer:45 degreesExplanation:Let&#039;s assume the mass of the block is m and the tension in the string is T.When the van is going straight ahead at a speed of 24 m/s, the block hangs vertically down, which means the tension in the string is equal to the weight of the block. We can express this as:T=mg(Equation&amp;nbsp;1)where g is the acceleration due to gravity.When the van maintains the same speed around the unbanked curve, the block swings toward the outside of the curve. At this point, there are two forces acting on the block: the tension in the string (T) and the centrifugal force (Fc).The centrifugal force is given by:Fc=mv2rwhere v is the speed of the van and r is the radius of the curve.Since the block is in equilibrium, the net force acting on it must be zero. The net force is the vector sum of the tension and the centrifugal force. Considering the horizontal and vertical components, we can write:Horizontal component:T=Fccos(θ)(Equation&amp;nbsp;2)Vertical component:mg=Fcsin(θ)(Equation&amp;nbsp;3)Dividing Equation 2 by Equation 3, we get:Tmg=Fccos(θ)Fcsin(θ)Simplifying further:tan(θ)=TmgFrom Equation 1, we know that T=mg, so:tan(θ)=Tmg=mgmg=1Therefore, θ=arctan(1).Substituting the value, we find:θ=arctan(1)=π4Hence, the angle θ is π4 or 45 degrees.In summary, the angle θ is 45 degrees.

Don Sumner · Accepted Answer

To solve this problem, we&#039;ll consider the forces acting on the block in both situations: when the van moves straight ahead and when it moves around the unbanked curve.When the van moves straight ahead at a constant speed, the block hangs vertically down, which means the tension in the string balances the force of gravity acting on the block. The force diagram in this case is:∑Fy=T−mg=0where T is the tension in the string and m is the mass of the block. Since the block is in equilibrium, the tension in the string is equal to the weight of the block:T=mgNow, when the van maintains the same speed around the unbanked curve, there is an additional force acting on the block due to its circular motion. This force is called the centripetal force and is given by:Fc=mv2rwhere v is the speed of the van and r is the radius of the curve.In this situation, the forces acting on the block can be resolved into two components: the vertical component, which is balanced by the tension in the string, and the horizontal component, which provides the centripetal force. The force diagram in this case is:∑Fy=T−mg=0∑Fx=Fc=mv2rSolving the first equation for T gives:T=mgSubstituting this into the second equation, we have:mg=mv2rCanceling out the mass m, we get:g=v2rFinally, we can find the angle θ by considering the triangle formed by the tension in the string, the vertical component of the weight, and the horizontal component of the centripetal force. This triangle is a right triangle, and we can write:tan(θ)=Fcmg=v2rgHence, the angle θ is given by:θ=tan−1(v2rg)Substituting the given values, with v=24m/s and r=175m, we can calculate θ as:θ=tan−1((24m/s)2(175m)(9.8m/s2))

RizerMix · Accepted Answer

Step 1:When the van moves straight ahead, the block hangs vertically down, indicating that the gravitational force acting downward is balanced by the tension in the string. Therefore, we have:Tstraight=mgwhere Tstraight is the tension in the string, m is the mass of the block, and g is the acceleration due to gravity.When the van maintains the same speed around the unbanked curve, an additional force acts on the block, known as the centrifugal force. This force is directed outward and causes the block to swing toward the outside of the curve.The centrifugal force is given by:Fcentrifugal=mv2rwhere v is the speed of the van, and r is the radius of the curve.In this scenario, the vertical component of the tension in the string must balance the gravitational force, while the horizontal component of the tension must balance the centrifugal force. Therefore, we can write:Tcurve,&amp;nbsp;vertical=mgTcurve,&amp;nbsp;horizontal=mv2rThe angle θ that the string makes with the vertical can be determined using the relationship:tanθ=Tcurve,&amp;nbsp;horizontalTcurve,&amp;nbsp;verticalStep 2:Substituting the values for Tcurve,&amp;nbsp;vertical and Tcurve,&amp;nbsp;horizontal, we have:tanθ=mv2rmgSimplifying, we get:tanθ=v2rgNow we can substitute the given values: v=24m/s and r=175m, and the acceleration due to gravity g≈9.8m/s2, into the equation for tanθ:tanθ=(24m/s)2(175m)(9.8m/s2)Evaluating the expression on the right-hand side gives us the value of tanθ. To find the angle θ, we can take the inverse tangent (or arctan) of this value:θ=arctan((24m/s)2(175m)(9.8m/s2))Evaluating this expression will give us the value of θ.

A block is hung by a string from inside the roof of avan. When the van goes straight ahead at a speed of 24 m/s,the block hangs vertically down. But w

Answered question

Answer & Explanation

New Questions in Force, Motion and Energy