SOLUTIONA delivery company has just unloaded a 500 N crate full of home exercise equipment inyour level driveway. Suppose you try to move the crate by pulling upward on the rope at anangle of 30° above the horizontal. How hard do you have to pull to keep the crate movingwith constant velocity? Is this easier or harder than pulling horizontally? For this crate thecoefficient of kinetic friction, u4, is 0.40.SET UP (Figure 1) shows the situation and our free-body diagram. The forces acting on the crate are its weight (magnitude w), the force applied by the rope (magnitude T), the normal (magnitude n), and the kineticfrictional component (magnitude fk) of the contact force the driveway surface exerts on the crate. The magnitude fk of the friction force is equal to un, but n is not equal in magnitude to the weight of the crate.Instead, the force exerted by the rope has an additional vertical component that tends to lift the crate off the floor, this component makes n less than w.SOLVE The crate is moving with constant velocity, so it is in equilibrium. Applying EF = 0 in component form, we findEF, =T cos 30° + (-fx) = T cos 30° – 0.40n = 0EF, = T sin 30° +n+ (-500 N) = 0These are two simultaneous equations for the two unknown quantities T and n. To solve them, we can eliminate one unknown and solve for the other. There are many ways to do this. Here is one way: Rearrange thesecond equation to the formn = 500 N - T sin 30°Then substitute this expression for n back into the first equation:T cos 30° – 0.40(500 N –T sin 30°) = 0Finally, solve this equation for T, and then substitute the result back into either of the original equations to obtain n. The results areT= 188 N,n= 406 NREFLECT The normal-force magnitude n is less than the weight (w = 500 N) of the box, and the tension required is a little less than the force that was needed (200 N) when you pulled horizontallyPart A - Practice Problem:Figure< 1 of 1If the coefficient of static friction is 0.46, how hard do you have to pull on the crate initially to get it moving if you pull upward on the rope at an angle of 20 ° above the horizontal?Express your answer in newtons.-T sin 30°30°fk = 0.40n'T =NT cos 30°30°W = 500 NSubmitPrevious Answers Request AnswerX Incorrect; Try Again; 3 attempts remaining

Name: SOLUTIONA delivery company has just unloaded a 500 N crate full of ho
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Description: SOLUTIONA delivery company has just unloaded a 500 N crate full of home exercise equipment inyour level driveway. Suppose you try to move the crate by pulling upward on the rope at anangle of 30° above the horizontal. How hard do you have to pull to

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SOLUTION A delivery company has just unloaded a 500 N crate full of home exercise equipment in your level driveway. Suppose you try to move the crate by pulling upward on the rope at an angle of 30 above the horizontal. How hard do you have to pull to keep the crate moving with constant velocity? Is this easier or harder than pulling horizontally? For this crate the coefficient of kinetic friction, . is 0.40. SET UP (Eiguro 1) shows the situation and our free-body diagram. The forces acting on the crate are its weight (magnitude w), the force applied by the rope (magnitude T), the normal (magnitude n), and the kinetic frictional component (magnitude f) of the contact force the driveway surface exerts on the crate. The magnitude fi of the friction force is equal to an, but n is not equal in magnitude to the weight of the crate Instead, the force exerted by the rope has an additional vertical component that tends to lift the crate off the floor, this component makes n less than tw SOLVE The crate is moving with constant velocity, so it is in equilibrium. Applying EF = 0 in component form, we find EF, =T cos 30 +(-) =T cos 30- 0.40n = 0 EF, =T sin 30° +n+ (-500 N) =0 These are two simultaneous equations for the two unknown quantities T and n. To solve them, we can eliminate one unknown and solve for the other. There are many ways to do this. Here is one way: Rearrange the second equation to the form n= 500 N- T sin 30 Then substitute this expression for rn back into the first equation T cos 30-0.40(500 N-T sin 30")-0 Finally, solve this equation for T, and then substitute the result back into either of the original equations to obtain n. The results are T- 188 N, n= 406 N REFLECT The normal-force magnitude n is less than the weight (w= 500 N) of the box, and the tension required is a little less than the force that was needed (200 N) when you pulled horizontally Figure < 1 of 1 • Part A - Practice Problem: If the coefficient of static friction 4, is 0.46, how hard do you have to pull on the crate initially to get it moving if you pull upward on the rope at an angle of 20" above the horizontal? Express your answer in newtons. sin 30 -0,40n 30° Tcos 30 T- 2959 w-500 N Submit Previous Answers Reauest Answer X Incorrect; Try Again; 3 attempts remaining