Question Video: Figuring Out the Weight and Normal Reaction of a Point Mass on a Slope Physics • 9th Grade

A point mass ๐‘ƒ is on a slope, as in the diagram. The weight of the mass, ๐‘Š, and the normal reaction force, ๐‘…, act on the point mass. What is the relationship between ๐‘… and ๐‘Š if the angle of the slope above the horizontal is zero? What is the magnitude of ๐‘… if the angle of the slope above the horizontal is 90ยฐ?

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Video Transcript

A point mass ๐‘ƒ is on a slope, as shown in the diagram. The weight of the mass, ๐‘Š, and the normal reaction force, ๐‘…, act on the point mass. What is the relationship between ๐‘… and ๐‘Š if the angle of the slope above the horizontal is zero? What is the magnitude of ๐‘… if the angle of the slope above the horizontal is 90 degrees?

In the first question, it asked us to find the relationship between ๐‘… and ๐‘Š, where ๐‘… is the normal reaction force and ๐‘Š is the weight of the object when the angle of the slope above the horizontal is zero. Letโ€™s redraw our diagram where we have an angle of zero. If the angle is zero degrees, that implies that we have a horizontal surface. Applying Newtonโ€™s second law to the situation using only the vertical forces as those are the only forces acting on our point, we get that ๐‘…, the normal reaction force, minus ๐‘Š, the weight, is equal to zero, as our point mass is not accelerating up and down.

We use the common convention that up is positive, thatโ€™s why that ๐‘… has a positive value, and that down is negative, which is why the weight ๐‘Š has a negative value. This means the magnitude of the normal reaction force is equal to the magnitude of the weight. Therefore, the relationship between ๐‘… and ๐‘Š is that they are equal. For the second question, we are asked to find the magnitude of ๐‘… if the angle of the slope above the horizontal is 90 degrees. Letโ€™s once again draw our diagram with an angle of 90 degrees this time.

To solve for this question, letโ€™s find a relationship for ๐‘… in terms of ๐œƒ. If we align our coordinate system to be parallel and perpendicular to the plane, then we must break down our weight vector into parallel and perpendicular components. We can see the normal reaction force will have the same magnitude as the perpendicular component of the weight. To find the perpendicular component of the weight, we can use trigonometry as the perpendicular component is the adjacent side of a right triangle to the angle ๐œƒ and the weight ๐‘Š is the hypotenuse of the triangle. So we can use the equation the cos of the angle ๐œƒ is equal to the adjacent side of a triangle divided by the hypotenuse of the triangle. Plugging in our values, we have cos ๐œƒ is equal to the perpendicular component of the weight divided by the weight.

To isolate the perpendicular component of the weight, we multiply both sides by ๐‘Š, canceling out the ๐‘Š on the right side of the equation, leaving us with the weight of the object times the cos of the angle ๐œƒ is equal to the perpendicular component of the weight. We said earlier that the magnitude of ๐‘… will be equal to the magnitude of the perpendicular component of the weight, so we can replace the perpendicular component with ๐‘…. We should also recall that weight is ๐‘š๐‘”, so we could replace the weight ๐‘Š with ๐‘š๐‘”. So the expression for the normal reaction force is ๐‘š๐‘” cos ๐œƒ. The cos of 90 degrees is zero, which means that the normal reaction force of point mass ๐‘ƒ when the angle of the slope is 90 degrees is zero. The magnitude of ๐‘… as the angle of the slope above the horizontal becomes 90 degrees is zero.

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